Write an equation for each parabola

Question

A parabola with x-intercepts at (-1,0) and (3,0) which passes through the point (1,-8)

A parabola with x-intercepts at 0 and 1 and which passes through the point (2,-2)

Answer 1

Hi Destiny B

Per my comment above regarding your parabola with x intercepts of 0 and 1 which passes through the point (2,-2)

In

f(x) = ax² + bx + c

c = 0 because of the point (0,0) so

f(x) = ax² + bx can easily be converted into two equations in two unknowns using the other two points

(1, 0) and (2, -2)

a(1)² + b(1) = 0

a(2)² + b(2) = -2

To give

a + b = 0

4a + 2b = -2

Eliminate b

-2a - 2b =0

4a + 2b = -2

Leaving

2a = -2

Dividing both sides by 2

a = -1

Solve for b

-1 + b = 0

b =1

Plugging these values into f(x) = ax² + bx to give

f(x) = -x² + x

As your parabola.

Of course, you can and should use a graphing calculator, desmos.com or another online calculator to see your parabolas and check your values for a, b, and c but you can also check them algebraically. Since you are given 3 points along your parabolas, you can try solving as 3 equations in 3 unknowns a, b, and c, for your first parabola, two equations in two unknowns for your second, and finally checking to see that the following is true:

ax² + bx + c = a(x - h)² + k

Standard Form and Vertex form should be equivalent and yield the same parabola

For your parabola passing through (-1,0), (3, 0) and (1, -8)

2x² - 4x -6 = 2(x - 1)² - 8

I hope this helps especially if you don’t have a graphing calculator or access to Desmos or you don’t know how to use either well enough yet.

Answer 2

Answer 3

y = a(x-h)² + k where (h,k) represents the vertex of the parabola.

Since the x-intercepts are at (-1, 0) and (3, 0), you know that the axis of symmetry (the line of symmetry) is the average of these x-values:

h = (-1 + 3)/2 = 1

So, y = a(x-1)² + k

Let's use (1,-8) to find

-8 = a. (0) + k ==> k = -8

y = a(x-1)² + 8, let's plug (-1,0)

Solve for a, a = 2

Finally the equation is y = 2(x-1)² - 8