FIND THE EQUATION OF QUADRATIC FUNCTION FROM A GRAPH

Determine the equation of quadratic function from graph. Give the function in general form.

Example 1 :

Solution :

From the graph, x-intercepts are 1 and 4.

So, x = 1 and x = 4

Factored form :

y = a(x - 1)(x - 4)

By observing the graph, the parabola cuts the y-axis at -4. By writing it as point (0, -4).

-4 = a(0 - 1)(0 - 4)

-4 = a(-1)(-4)

-4 = 4a

a = -1

y = -1(x - 1)(x - 4)

y = -1(x² - 5x + 4)

y = -x² + 5x - 4

Here the coefficient of x² is -1, from the graph we have evidence that the parabola opens down.

Example 2 :

Solution :

From the graph, x-intercepts are 1 and 4.

So, x = 1 and x = 4

Factored form :

y = a(x - 1)(x - 4)

By observing the graph, the parabola cuts the y-axis at -4. By writing it as point (0, 4).

4 = a(0 - 1)(0 - 4)

4 = a(-1)(-4)

4 = 4a

a = 1

y = 1(x - 1)(x - 4)

y = 1(x² - 5x + 4)

y = x² - 5x + 4

Here the coefficient of x² is 1, from the graph we have evidence that the parabola opens up.

Example 3 :

Solution :

From the graph, x-intercepts are -4 and 1.

So, x = -4 and x = 1

Factored form :

y = a(x + 4)(x - 1)

By observing the graph, the parabola cuts the y-axis at 12. By writing it as point (0, 12).

12 = a(0 + 4)(0 - 1)

12 = -4a

a = -3

y = -3(x + 4)(x - 1)

y = -3(x² + 3x - 4)

y = -3x² - 9x + 12

Here the coefficient of x² is -3, from the graph we have evidence that the parabola opens down.

Example 4 :

Solution :

From the graph, x-intercepts are -4 and 1.

So, x = -4 and x = 1

Factored form :

y = a(x + 4)(x - 1)

By observing the graph, the parabola cuts the y-axis at -8. By writing it as point (0, -8).

-8 = a(0 + 4)(0 - 1)

-8 = -4a

a = 2

y = 2(x + 4)(x - 1)

y = 2(x² + 3x - 4)

y = 2x² + 6x - 8

Here the coefficient of x² is 2, from the graph we have evidence that the parabola opens up.

Example 5 :

Solution :

From the graph, x-intercepts are 4 and 1.

So, x = 4 and x = 1

Factored form :

y = a(x - 4)(x - 1)

By observing the graph, the parabola cuts the y-axis at -8. By writing it as point (0, 8).

8 = a(0 - 4)(0 - 1)

8 = 4a

a = 2

y = 2(x - 4)(x - 1)

y = 2(x² - 5x + 4)

y = 2x² - 10x + 8

Here the coefficient of x² is 2, from the graph we have evidence that the parabola opens up.

Example 6 :

The area of a rectangle is modeled by the graph where y is the area (in square meters) and x is the width (in meters). Write an equation of the parabola. Find the dimensions and corresponding area of one possible rectangle. What dimensions result in the maximum area?

Solution :

By observing the graph, the x-intercepts are 0 and 7

The factored form is (x - 0)(x - 7)

f(x) = a(x)(x - 7)

The parabola passes through the point (1, 6), then

f(1) = a(1)(1 - 7)

6 = a(-6)

a = -1

Applying the value of a, we get

f(x) = -x(x - 7)

Example 7 :

Every rope has a safe working load. A rope should not be used to lift a weight greater than its safe working load. The table shows the safe working loads S (in pounds) for ropes with circumference C (in inches). Write an equation for the safe working load for a rope. Find the safe working load for a rope that has a circumference of 10 inches.

Solution :

y = a(x - p)(x - q)

Here x -intercept is 0.

y = a(x - 0)(x - q)

y = a(x)(x - q)

The parabola passes through the points (1, 180) and (2, 720)

180 = a(1)(1 - q)

180 = a(1 - q)

a = 180 / (1 - q) -----(1)

720 = a(2)(2 - q)

720 = 2a(2 - q)

360 = a(2 - q)

a = 360 / (2 - q) -----(2)

(1) = (2)

180 / (1 - q) = 360 / (2 - q)

(2 - q)/(1 - q) = 360/180

(2 - q)/(1 - q) = 2

2 - q = 2(1 - q)

2 - q = 2 - 2q

2 - 2 = -2q + q

q = 0

Applying the value of q, we get

a = 180 / (1 - 0)

a = 180

Applying these values, we get

y = 180(x)(x - 0)

y = 180x²