Find the value of h(1)

Question

Suppose the graph of y=h(x) is a parabola with axis of symmetry x=5, with h(5)=0 and h(9)=10. 

Find the value of h(1).

Answer 1

The axis of symmetry is also the x-coordinate of the parabola's vertex.  Since we are given h(5)=0, we know it's vertex.  A parabola in vertex form is

 

y = a(x - b)² + k

 

where:

a is the coefficient of x²

(b, k) is the vertex

 

 

The vertex is (5, 0) and is also the x-intercept because the y-coordinate is zero.  So plugging this into the vertex form,

 

y = a(x - 5)²

 

 

Next, we need to solve for a.  We use the fact that   h(9)=10.  Plugging in these values into the vertex form,

 

10 = a(9 - 5)²

 

10 = 4²a

 

10 = 16a

 

5/8 = a

 

 

So your actual equation of the parabola in vertex form is

 

h(x) = (5/8)(x - 5)²

 

This is a parabola that open upward.  So the vertex is a minimum value.

 

 

To find h(1), evaluate this function when x=1.

 

 

Answer 2

A parabola has a shape of a "U" (or an upside down "U".)

The axis of symmetry is where the 'middle' or 'center' of the "U" is,

this is also where the lowest point of the "U" is.

Making a little table, here is what we know and don't know

for this parabola:

 

x   h(x)

 

1    ?

 

5    0    <-- axis of symmetry

 

9   10

 

h(x) is a "U" shape and x=5 is the 'center' of the "U".

x=9 is 4 units greater than the center and has a height h(x)=10

So, at x=1, which is 4 units less than the center,

we'd also expect a height of 10, h(1)  = 10.

 

Drawing a sketch would be better to show this ;-)