Question

Using the vertex (0.5,9)
Axis of symmetry x=(0.5)
y intercept (0,5)
x intercepts (0,0), (1.25,0)
Using h(t) =
Write as an equation

Answer 1

To write the equation of a parabola given the vertex, axis of symmetry, y-intercept, and x-intercepts, we can use the vertex form of a quadratic equation:

\[
h(t) = a(t - h)^2 + k
\]

where \((h, k)\) is the vertex of the parabola.

Given the vertex \((0.5, 9)\), we have:

\[
h(t) = a(t - 0.5)^2 + 9
\]

Next, we need to determine the coefficient \(a\). We can use one of the intercepts to find \(a\). Let's use the y-intercept \((0, 5)\):

\[
h(0) = a(0 - 0.5)^2 + 9
\]
\[
5 = a(0.5)^2 + 9
\]
\[
5 = a(0.25) + 9
\]
\[
5 - 9 = 0.25a
\]
\[
-4 = 0.25a
\]
\[
a = -4 / 0.25 = -16
\]

Now we have determined \(a = -16\). Thus, the equation becomes:

\[
h(t) = -16(t - 0.5)^2 + 9
\]

This is the equation of the parabola with the given properties. If desired, we can expand this into standard form:

1. Expand the square:
\[
(t - 0.5)^2 = t^2 - t + 0.25
\]
2. Substitute it back into the equation:
\[
h(t) = -16(t^2 - t + 0.25) + 9
\]
3. Distribute \( -16 \):
\[
h(t) = -16t^2 + 16t - 4 + 9
\]
4. Combine like terms:
\[
h(t) = -16t^2 + 16t + 5
\]

So the equation in standard form is:

\[
h(t) = -16t^2 + 16t + 5
\]