Graph of Quadratic Polynomial
The graph of a polynomial provides a visual representation of how the polynomial behaves for different values of the variable. For quadratic polynomials of the form ax^2 + bx + c, the graph is always a smooth U-shaped curve called a parabola. Understanding the graph of a quadratic polynomial helps us visualise its zeroes (where the graph crosses the x-axis), determine whether it opens upward or downward, and locate its vertex (the turning point). In Class 10, we focus on sketching the graph of quadratic polynomials, reading the zeroes from the graph, and understanding how the sign of the leading coefficient affects the shape of the parabola. This graphical understanding complements the algebraic methods of finding zeroes and connects algebra with coordinate geometry in a meaningful way.
What is Graph of a Quadratic Polynomial?
A quadratic polynomial is a polynomial of degree 2, written in the standard form p(x) = ax^2 + bx + c, where a, b, and c are real numbers and a is not equal to zero. The graph of this polynomial is obtained by plotting the points (x, p(x)) for various values of x on the Cartesian plane and joining them with a smooth curve.
Parabola: The graph of every quadratic polynomial is a curve called a parabola. A parabola is symmetric about a vertical line called the axis of symmetry. The point where the parabola turns (its lowest or highest point) is called the vertex.
Direction of Opening:
- If a > 0 (positive leading coefficient), the parabola opens upward (U-shape). The vertex is the minimum point.
- If a < 0 (negative leading coefficient), the parabola opens downward (inverted U-shape). The vertex is the maximum point.
Vertex: The vertex of the parabola y = ax^2 + bx + c is at the point (-b/(2a), p(-b/(2a))). The x-coordinate of the vertex is -b/(2a), and the y-coordinate is obtained by substituting this value into the polynomial.
Axis of Symmetry: The vertical line x = -b/(2a) passes through the vertex and divides the parabola into two mirror-image halves.
Zeroes from the Graph: The zeroes of the polynomial are the x-coordinates of the points where the parabola intersects the x-axis. A quadratic polynomial can have 0, 1, or 2 real zeroes, depending on whether the parabola does not touch the x-axis, touches it at exactly one point, or crosses it at two points.
Graph of Quadratic Polynomial Formula
Standard form: y = ax^2 + bx + c
Vertex: (-b/(2a), c - b^2/(4a))
Axis of symmetry: x = -b/(2a)
Discriminant: D = b^2 - 4ac
If D > 0: The parabola crosses the x-axis at two distinct points (two real zeroes).
If D = 0: The parabola touches the x-axis at exactly one point (one repeated zero, vertex on x-axis).
If D < 0: The parabola does not intersect the x-axis (no real zeroes).
Zeroes (x-intercepts): x = (-b + sqrt(D))/(2a) and x = (-b - sqrt(D))/(2a), when D >= 0.
Y-intercept: The graph crosses the y-axis at the point (0, c).
Derivation and Proof
The shape of the graph of a quadratic polynomial can be understood by completing the square.
Step 1: Start with y = ax^2 + bx + c.
Step 2: Factor out 'a' from the first two terms: y = a(x^2 + (b/a)x) + c.
Step 3: Complete the square inside the parentheses. To complete the square for x^2 + (b/a)x, add and subtract (b/(2a))^2 = b^2/(4a^2):
y = a[x^2 + (b/a)x + b^2/(4a^2) - b^2/(4a^2)] + c
= a[(x + b/(2a))^2 - b^2/(4a^2)] + c
= a(x + b/(2a))^2 - b^2/(4a) + c
= a(x + b/(2a))^2 + (4ac - b^2)/(4a)
Step 4: This is now in vertex form: y = a(x - h)^2 + k, where h = -b/(2a) and k = (4ac - b^2)/(4a).
Why this gives a parabola: The expression a(x - h)^2 + k shows that y is a squared function of (x - h), shifted by h horizontally and k vertically. The squared term (x - h)^2 is always non-negative, so:
If a > 0: y >= k for all x, with minimum value k at x = h. The parabola opens upward.
If a < 0: y <= k for all x, with maximum value k at x = h. The parabola opens downward.
Why the parabola is symmetric: Notice that (x - h)^2 = (-(x - h))^2. This means points equidistant from x = h on either side give the same y-value. Hence the parabola is symmetric about the line x = h.
Effect of the coefficient 'a': The magnitude of 'a' determines how wide or narrow the parabola is. A larger |a| makes the parabola narrower (steeper), while a smaller |a| makes it wider (flatter). The sign of 'a' determines the direction of opening.
Connection to zeroes: Setting y = 0 gives a(x - h)^2 + k = 0, or (x - h)^2 = -k/a. This has real solutions only when -k/a >= 0. Substituting k = (4ac - b^2)/(4a) and simplifying leads to the condition b^2 - 4ac >= 0, which is the discriminant condition.
Types and Properties
The graph of a quadratic polynomial can take several forms depending on the coefficients and the discriminant:
Type 1: Upward-opening parabola with two x-intercepts (a > 0, D > 0)
Example: y = x^2 - 5x + 6. Here a = 1 > 0 and D = 25 - 24 = 1 > 0. The graph opens upward and crosses the x-axis at x = 2 and x = 3. The vertex is below the x-axis.
Type 2: Upward-opening parabola with one x-intercept (a > 0, D = 0)
Example: y = x^2 - 4x + 4 = (x - 2)^2. The graph touches the x-axis at x = 2 (the vertex lies on the x-axis). Both zeroes are equal.
Type 3: Upward-opening parabola with no x-intercept (a > 0, D < 0)
Example: y = x^2 + x + 1. Here D = 1 - 4 = -3 < 0. The entire graph lies above the x-axis. There are no real zeroes.
Type 4: Downward-opening parabola with two x-intercepts (a < 0, D > 0)
Example: y = -x^2 + 4x - 3. Here a = -1 < 0 and D = 16 - 12 = 4 > 0. The graph opens downward and crosses the x-axis at x = 1 and x = 3. The vertex is above the x-axis.
Type 5: Downward-opening parabola with one x-intercept (a < 0, D = 0)
Example: y = -x^2 + 6x - 9 = -(x - 3)^2. The vertex is on the x-axis at x = 3, and the parabola opens downward.
Type 6: Downward-opening parabola with no x-intercept (a < 0, D < 0)
Example: y = -x^2 - 2x - 5. Here D = 4 - 20 = -16 < 0. The entire graph lies below the x-axis. No real zeroes.
Methods
Method: Steps to Sketch the Graph of a Quadratic Polynomial
Step 1: Identify a, b, and c from the polynomial y = ax^2 + bx + c.
Step 2: Determine the direction of opening. If a > 0, the parabola opens upward. If a < 0, it opens downward.
Step 3: Find the vertex. The x-coordinate of the vertex is h = -b/(2a). Substitute h into the polynomial to find the y-coordinate: k = p(h).
Step 4: Find the y-intercept by setting x = 0: y-intercept = (0, c).
Step 5: Find the x-intercepts (zeroes) by solving ax^2 + bx + c = 0. Calculate D = b^2 - 4ac. If D >= 0, the zeroes are x = (-b + sqrt(D))/(2a) and x = (-b - sqrt(D))/(2a).
Step 6: Plot the vertex, y-intercept, x-intercepts (if any), and a few additional points on either side of the vertex.
Step 7: Draw a smooth parabolic curve through these points, ensuring symmetry about the axis of symmetry x = -b/(2a).
Reading Zeroes from a Graph: To find the zeroes from a given graph, identify the x-coordinates where the curve crosses or touches the x-axis. These x-values are the zeroes of the polynomial.
Solved Examples
Example 1: Example 1: Sketch the graph of y = x^2 - 4x + 3
Problem: Draw the graph of y = x^2 - 4x + 3 and find its zeroes.
Solution:
Here a = 1, b = -4, c = 3.
Direction: a = 1 > 0, so the parabola opens upward.
Vertex: x = -b/(2a) = 4/2 = 2. y = (2)^2 - 4(2) + 3 = 4 - 8 + 3 = -1. Vertex = (2, -1).
Y-intercept: (0, 3).
X-intercepts: x^2 - 4x + 3 = 0 gives (x - 1)(x - 3) = 0, so x = 1 and x = 3.
Table of values: x = -1: y = 1 + 4 + 3 = 8. x = 0: y = 3. x = 1: y = 0. x = 2: y = -1. x = 3: y = 0. x = 4: y = 3. x = 5: y = 8.
Plot these points and draw a smooth U-shaped curve. The graph crosses the x-axis at (1, 0) and (3, 0).
Answer: The zeroes are x = 1 and x = 3.
Example 2: Example 2: Sketch the graph of y = -x^2 + 2x + 3
Problem: Sketch the graph of y = -x^2 + 2x + 3 and identify its zeroes, vertex, and axis of symmetry.
Solution:
Here a = -1, b = 2, c = 3.
Direction: a = -1 < 0, so the parabola opens downward.
Vertex: x = -b/(2a) = -2/(2(-1)) = 1. y = -(1)^2 + 2(1) + 3 = -1 + 2 + 3 = 4. Vertex = (1, 4).
Axis of symmetry: x = 1.
Y-intercept: (0, 3).
X-intercepts: -x^2 + 2x + 3 = 0. Multiply by -1: x^2 - 2x - 3 = 0, giving (x - 3)(x + 1) = 0. So x = 3 and x = -1.
Table: x = -2: y = -4 - 4 + 3 = -5. x = -1: y = 0. x = 0: y = 3. x = 1: y = 4. x = 2: y = 3. x = 3: y = 0. x = 4: y = -5.
The graph is an inverted U-shape with the highest point at (1, 4), crossing the x-axis at (-1, 0) and (3, 0).
Answer: Zeroes: x = -1 and x = 3. Vertex: (1, 4). Axis: x = 1.
Example 3: Example 3: Graph of a perfect square quadratic
Problem: Sketch the graph of y = x^2 - 6x + 9 and comment on the number of zeroes.
Solution:
a = 1, b = -6, c = 9. Note that x^2 - 6x + 9 = (x - 3)^2.
Direction: Opens upward (a > 0).
Vertex: x = -(-6)/(2*1) = 3. y = (3-3)^2 = 0. Vertex = (3, 0).
D = b^2 - 4ac = 36 - 36 = 0. Exactly one zero (repeated).
Table: x = 0: y = 9. x = 1: y = 4. x = 2: y = 1. x = 3: y = 0. x = 4: y = 1. x = 5: y = 4. x = 6: y = 9.
The parabola touches the x-axis at exactly one point (3, 0) and lies entirely above the x-axis otherwise. The vertex sits on the x-axis.
Answer: There is exactly one zero: x = 3 (repeated). The parabola just touches the x-axis without crossing it.
Example 4: Example 4: Graph with no real zeroes
Problem: Sketch the graph of y = x^2 + 2x + 3 and determine the number of zeroes.
Solution:
a = 1, b = 2, c = 3.
Direction: Opens upward (a > 0).
Vertex: x = -2/(2*1) = -1. y = 1 - 2 + 3 = 2. Vertex = (-1, 2).
D = 4 - 12 = -8 < 0. No real zeroes.
Y-intercept: (0, 3).
Table: x = -3: y = 9 - 6 + 3 = 6. x = -2: y = 4 - 4 + 3 = 3. x = -1: y = 2. x = 0: y = 3. x = 1: y = 6.
The entire parabola lies above the x-axis. The minimum point is (−1, 2), which is above the x-axis, so the curve never crosses or touches the x-axis.
Answer: The polynomial has no real zeroes. The graph does not intersect the x-axis.
Example 5: Example 5: Determine zeroes from a described graph
Problem: The graph of a quadratic polynomial opens upward, has its vertex at (2, -9), and crosses the x-axis at x = -1 and x = 5. Find the polynomial.
Solution:
Since the zeroes are -1 and 5, we can write the polynomial as: p(x) = a(x - (-1))(x - 5) = a(x + 1)(x - 5).
The graph opens upward, so a > 0.
The vertex is at x = (-1 + 5)/2 = 2 (midpoint of zeroes). At x = 2: p(2) = a(3)(-3) = -9a. We are told the vertex y-coordinate is -9, so -9a = -9, giving a = 1.
Therefore, p(x) = (x + 1)(x - 5) = x^2 - 4x - 5.
Verification: Vertex x = -(-4)/(2*1) = 2. Vertex y = 4 - 8 - 5 = -9. Zeroes: x^2 - 4x - 5 = (x + 1)(x - 5) = 0 gives x = -1 and x = 5. All correct.
Answer: The polynomial is x^2 - 4x - 5.
Example 6: Example 6: How many zeroes does the graph reveal?
Problem: Without drawing the graph, determine how many zeroes the polynomial y = 2x^2 - 3x + 5 has by analysing the discriminant.
Solution:
a = 2, b = -3, c = 5.
D = b^2 - 4ac = 9 - 40 = -31.
Since D < 0, the quadratic equation 2x^2 - 3x + 5 = 0 has no real solutions.
Since a = 2 > 0, the parabola opens upward.
The vertex y-coordinate = c - b^2/(4a) = 5 - 9/8 = 31/8 > 0.
So the entire graph lies above the x-axis.
Answer: The polynomial has no real zeroes. The graph does not intersect the x-axis at any point.
Example 7: Example 7: Compare two parabolas
Problem: Compare the graphs of y = x^2 and y = 3x^2. How do they differ?
Solution:
Both have the same vertex at the origin (0, 0), open upward, and have x = 0 as the axis of symmetry.
Table for y = x^2: x = 1: y = 1. x = 2: y = 4. x = -1: y = 1.
Table for y = 3x^2: x = 1: y = 3. x = 2: y = 12. x = -1: y = 3.
For the same x-value, y = 3x^2 gives a larger y-value (3 times as large). This means y = 3x^2 rises more steeply. The parabola y = 3x^2 is narrower than y = x^2.
General rule: A larger value of |a| makes the parabola narrower (steeper), while a smaller value makes it wider (flatter). The sign of a determines the direction.
Answer: Both open upward with vertex at origin, but y = 3x^2 is narrower (steeper) than y = x^2 because |3| > |1|.
Example 8: Example 8: Sketch y = -2x^2 + 8x - 6
Problem: Sketch the graph of y = -2x^2 + 8x - 6 and find the zeroes, vertex, and the maximum value of y.
Solution:
a = -2, b = 8, c = -6.
Direction: a < 0, opens downward.
Vertex: x = -8/(2*(-2)) = -8/(-4) = 2. y = -2(4) + 8(2) - 6 = -8 + 16 - 6 = 2. Vertex = (2, 2).
Maximum value of y = 2 (since the parabola opens downward, the vertex is the highest point).
Y-intercept: (0, -6).
X-intercepts: -2x^2 + 8x - 6 = 0. Divide by -2: x^2 - 4x + 3 = 0, so (x - 1)(x - 3) = 0. Zeroes: x = 1 and x = 3.
Table: x = -1: y = -2 - 8 - 6 = -16. x = 0: y = -6. x = 1: y = 0. x = 2: y = 2. x = 3: y = 0. x = 4: y = -6.
Answer: Zeroes: x = 1 and x = 3. Vertex: (2, 2). Maximum y = 2.
Example 9: Example 9: Determine the quadratic from a graph description
Problem: A parabola opens downward, has vertex at (-1, 8), and passes through the origin. Find the quadratic polynomial.
Solution:
Using vertex form: y = a(x - h)^2 + k = a(x - (-1))^2 + 8 = a(x + 1)^2 + 8.
The parabola passes through (0, 0): 0 = a(0 + 1)^2 + 8 = a + 8. So a = -8.
y = -8(x + 1)^2 + 8 = -8(x^2 + 2x + 1) + 8 = -8x^2 - 16x - 8 + 8 = -8x^2 - 16x.
Verification: At x = -1: y = -8(1) + 16 = 8. Correct (vertex). At x = 0: y = 0. Correct. a = -8 < 0. Opens downward. Correct.
Zeroes: -8x^2 - 16x = -8x(x + 2) = 0, so x = 0 and x = -2.
Answer: The polynomial is -8x^2 - 16x (or equivalently, -8x(x + 2)).
Example 10: Example 10: Read zeroes from a table of values
Problem: The following values are given for a quadratic polynomial p(x): p(-2) = 5, p(-1) = 0, p(0) = -3, p(1) = -4, p(2) = -3, p(3) = 0, p(4) = 5. Find the zeroes and the polynomial.
Solution:
From the table, p(-1) = 0 and p(3) = 0. So the zeroes are x = -1 and x = 3.
The polynomial can be written as p(x) = a(x + 1)(x - 3).
Using p(0) = -3: a(0 + 1)(0 - 3) = -3a = -3, so a = 1.
p(x) = (x + 1)(x - 3) = x^2 - 2x - 3.
Verification: p(-2) = 4 + 4 - 3 = 5. p(1) = 1 - 2 - 3 = -4. p(2) = 4 - 4 - 3 = -3. p(4) = 16 - 8 - 3 = 5. All match.
Vertex: x = -(-2)/(2*1) = 1. p(1) = -4. Vertex = (1, -4). This is the minimum point.
Answer: Zeroes are x = -1 and x = 3. The polynomial is x^2 - 2x - 3.
Real-World Applications
The graph of a quadratic polynomial has numerous applications in mathematics, science, and everyday life.
Projectile Motion: When an object is thrown upward at an angle, its path (trajectory) is a parabola. The quadratic polynomial models the height as a function of time or horizontal distance. The maximum height corresponds to the vertex, and the zeroes represent the launch and landing points.
Optimisation Problems: Since the vertex of a parabola gives either the maximum or minimum value, quadratic functions are used to solve optimisation problems. For example, finding the maximum area that can be enclosed with a fixed perimeter, or the price that maximises revenue.
Architecture and Engineering: Parabolic shapes appear in bridges, satellite dishes, and headlight reflectors. The reflective property of parabolas (all rays parallel to the axis reflect through the focus) makes them ideal for these applications.
Economics: Revenue and profit functions are often quadratic. The graph helps businesses identify break-even points (zeroes) and maximum profit (vertex).
Sports: The trajectory of a ball in cricket, basketball, or football follows a parabolic path, and understanding the graph helps in analysing and predicting the ball's motion.
Key Points to Remember
- The graph of a quadratic polynomial y = ax^2 + bx + c is always a parabola.
- If a > 0, the parabola opens upward (U-shape); if a < 0, it opens downward (inverted U).
- The vertex (turning point) is at (-b/(2a), p(-b/(2a))).
- The axis of symmetry is the vertical line x = -b/(2a).
- The zeroes of the polynomial are the x-intercepts of the graph.
- The discriminant D = b^2 - 4ac determines the number of zeroes: D > 0 gives 2 zeroes, D = 0 gives 1 zero, D < 0 gives no real zeroes.
- The y-intercept is always (0, c).
- A larger |a| makes the parabola narrower; a smaller |a| makes it wider.
- The vertex gives the minimum value of p(x) when a > 0, and the maximum value when a < 0.
- The parabola is always symmetric about the axis of symmetry — points equidistant from the axis have equal y-values.
Practice Problems
- Sketch the graph of y = x^2 + 6x + 8 and find its zeroes, vertex, and axis of symmetry.
- Without drawing the graph, determine how many zeroes the polynomial y = -3x^2 + 5x - 4 has.
- The graph of a quadratic polynomial has vertex at (3, -4) and passes through (5, 0). Find the polynomial.
- Sketch y = x^2 - 1 and y = -x^2 + 1 on the same axes. How many points of intersection are there?
- A quadratic polynomial has zeroes at x = -3 and x = 7. Its graph passes through (0, 21). Find the polynomial and its vertex.
- For what value of k does the graph of y = x^2 - kx + 9 touch the x-axis at exactly one point?
Frequently Asked Questions
Q1. What shape is the graph of a quadratic polynomial?
The graph of a quadratic polynomial is always a parabola — a smooth, U-shaped curve. If the coefficient of x^2 is positive, the U opens upward. If negative, it opens downward (inverted U).
Q2. How do you find the zeroes from the graph?
The zeroes of a quadratic polynomial are the x-coordinates of the points where the graph (parabola) intersects the x-axis. If the parabola crosses the x-axis at two points, there are two zeroes. If it touches the x-axis at one point, there is one repeated zero. If it doesn't touch the x-axis, there are no real zeroes.
Q3. What is the vertex of a parabola?
The vertex is the turning point of the parabola — the lowest point if it opens upward, or the highest point if it opens downward. For y = ax^2 + bx + c, the vertex is at x = -b/(2a), and the y-coordinate is found by substituting this x-value into the polynomial.
Q4. What determines whether the parabola opens up or down?
The sign of the leading coefficient 'a' determines the direction. If a > 0 (positive), the parabola opens upward. If a < 0 (negative), it opens downward. The magnitude of 'a' affects the width — larger |a| means a narrower parabola.
Q5. Can a quadratic polynomial have more than 2 zeroes?
No. A quadratic polynomial (degree 2) can have at most 2 zeroes. This is because a parabola can intersect the x-axis at most at 2 points. In general, a polynomial of degree n can have at most n real zeroes.
Q6. What is the axis of symmetry?
The axis of symmetry is a vertical line that passes through the vertex and divides the parabola into two mirror-image halves. Its equation is x = -b/(2a). For any two points on the parabola at equal distances from the axis of symmetry, the y-values are the same.
Q7. What is the discriminant and how does it relate to the graph?
The discriminant D = b^2 - 4ac tells us how many times the parabola intersects the x-axis. If D > 0, the parabola crosses the x-axis at two distinct points (2 zeroes). If D = 0, it touches the x-axis at one point (1 repeated zero). If D < 0, it does not intersect the x-axis (no real zeroes).
Q8. How is the graph of y = ax^2 different from y = ax^2 + bx + c?
y = ax^2 is the simplest parabola with vertex at the origin (0, 0). Adding b shifts the vertex horizontally and diagonally, while adding c shifts it vertically. The general parabola y = ax^2 + bx + c has the same shape as y = ax^2 but is repositioned so its vertex is at (-b/(2a), c - b^2/(4a)).
Q9. Why is the graph of a linear polynomial not a parabola?
A linear polynomial y = ax + b (degree 1) produces a straight line, not a parabola. The parabolic shape comes specifically from the x^2 term. Only polynomials of degree 2 (quadratic) produce parabolas.
Q10. How do you find the maximum or minimum value of a quadratic polynomial from the graph?
The maximum or minimum value is the y-coordinate of the vertex. If a > 0, the vertex gives the minimum value (the lowest point of the U). If a < 0, the vertex gives the maximum value (the highest point of the inverted U). The value is c - b^2/(4a).
Related Topics
- Zeroes of Quadratic Polynomial
- Relationship Between Zeroes and Coefficients
- Graph of Cubic Polynomial
- Discriminant of Quadratic Equation
- Polynomials in One Variable
- Degree of a Polynomial
- Types of Polynomials
- Value of a Polynomial
- Zeroes of a Polynomial
- Remainder Theorem
- Factor Theorem
- Factorisation of Polynomials
- Algebraic Identities (Extended)
- (a + b)³ and (a - b)³ Identities










