Question

what is the axis of symmetry, x-intercept, and y-intercept the equation, h(t)=-16t^2+64t+80?

Answer 1

Given a parabola, it's easiest to find its axis of symmetry by writing it in vertex form. (Alternatively, if you're taking calculus, you can find the vertex in a different way.)

So your job would be to write a quadratic given in this form,
\[h(t)=y=at^2+bt+c,\]
to this form,
\[y=a(t-j)^2+k,\]
where \((j,k)\) is the vertex of the parabola, and hence the line \(x=j\) makes up the axis of symmetry.

To change from the first form to the second, you must complete the square.

Answer 2

So, given \(h(t)=y=-16t^2+64t+80\), you complete the square as follows:
\[y=-16\left(t^2-4t-5\right)\\
y=-16\left(t^2-4t+4-9\right)\\
y=-16\left((t-2)^2-9\right)\\
y=-16(t-2)^2+144\]
Thus, the vertex is \((2,144)\), which means the axis of symmetry is the line \(t=2\).

Answer 3

To find the x-intercept (or rather, t-intercept), you have to use the fact that any intercept on the horizontal axis has a y-coordinate of 0, i.e. is given by the coordinate pair \((t,0)\).

This means that for this \(t\), you get \(y=0\), so you would find this \(t\) by solving the equation,
\[0=-16(t-2)^2+144\]
Finding the y-intercept is similar, the only difference being that you find the value of \(y\) when \(t=0\):
\[y=-16(0-2)^2+144\]