# How to graph a parabola #h(t)=-16t^2+280t+17?

Jun 20, 2015

$h \left(t\right) = - 16 {t}^{2} + 280 t + 17$ is a very steep parabola with vertex at $\left(8.75 , 1242\right)$ and intersections with the $t$ axis at approximately $\left(- 0.06 , 0\right)$ and $\left(17.56 , 0\right)$

#### Explanation:

$h \left(t\right) = - 16 {t}^{2} + 280 t + 17$

In general, we have:

$a {t}^{2} + b t + c = a {\left(t + \frac{b}{2 a}\right)}^{2} + \left(c - {b}^{2} / \left(4 a\right)\right)$

The vertex of this parabola is $\left(- \frac{b}{2} a , c - {b}^{2} / \left(4 a\right)\right)$

The axis of symmetry is $t = - \frac{b}{2 a}$

The intersections (if any) with the $t$ axis are at:

$\left(\frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a} , 0\right)$

The intersection with the vertical axis is at:

$\left(0 , c\right)$

In our case, we have $a = - 16$, $b = 280$ and $c = 17$

The discriminant $\Delta$ is given by the formula:

$\Delta = {b}^{2} - 4 a c = {280}^{2} - \left(4 \times - 16 \times 17\right) = 79488$

$= {2}^{7} \cdot {3}^{3} \cdot 23 = {24}^{2} \cdot 138$

Being positive, but not a perfect square, the parabola does intersect the $t$ axis at the points:

$\left(\frac{280 \pm 24 \sqrt{138}}{32} , 0\right) = \left(\frac{35 \pm 3 \sqrt{138}}{4} , 0\right)$

That is approximately $\left(- 0.06 , 0\right)$ and $\left(17.56 , 0\right)$

The axis of symmetry is $x = \frac{35}{4}$.

The vertex is at $\left(\frac{35}{4} , 17 + {35}^{2}\right) = \left(8.75 , 1242\right)$

This is parabola is very 'steep', so it will help to have a very different scale on the horizontal and vertical axes.

Here I have graphed $h \left(\frac{x}{100}\right)$

graph{-16(x/100)^2+280(x/100)+17 [-2128, 3710, -1520, 1400]}