How do I find the vertex given the following information?

Question

y-intercept -6.75, x-intercept 1, equation of axis of symmetry x-5=0 

Answer 1

Lets put this equation in vertex form.

 

y = a(x - h)² + k

 

where:

a is the leading coefficient

(h, k) is the vertex

 

Since the axis of symmetry is     x - 5 = 0

 

y = a(x - 5)² + k

 

 

Next, we will set two equations.  Onc for each intercept.

 

y-intercept:

 

-6.75 = 25a + k

 

 

x-intercept:

 

0 = 16a + k

 

 

Here are your two equations:

 

-6.75 = 25a + k                   eq1

0 = 16a + k                          eq2

 

 

We can solve for a and  k from these equation.   From eq2 ,    k = -16a

 

 

Substitute this value of k into eq1.

 

-6.75 = 25a - 16a

 

 

Solving for a.

 

-6.75 = 9a

 

-0.75 = a

 

 

Then,  k = (-16)(-0.75) = 12

 

 

Your equation is then,

 

y = -0.75(x - 5)² + 12

 

 

Vertex is (5, 12)

Answer 2

You have one x-intercept at x = 1.  The axis of symmetry is at x = 5.  Therefore the other x-intercept is at an equal distance (4) on the other side of the axis of symmetry, at x = 9.

 

y = a(x-1)(x-9)

 

To find the value of a, plug in the y-intercept (0,-6.75):

 

-6.75 = a(0-1)(0-9)

-6.75 = 9a

-6.75/9 = -3/4 = a

 

y = -(3/4)(x-1)(x-9)

 

The x-coordinate of the vertex is x=5 (along the axis of symmetry).  The y-coordinate is:

 

y = (-3/4)(5-1)(5-9)

y = (-3/4)(4)(-4)

y = 12

 

The vertex is located at (5,12)