# Quadratic formula ?

Mar 14, 2018

Alex will land $3.9$ meters from the ramp.

#### Explanation:

We seek to find $h \left(d\right) = 0$.

$0 = - 3.9 {d}^{2} + 13.1 d + 8.7$

Now apply the quadratic formula.

$d = \frac{- 13.1 \pm \sqrt{{13.1}^{2} - 4 \cdot - 3.9 \cdot 8.7}}{2 \cdot - 3.9}$

$d = \frac{- 13.1 \pm \sqrt{307.33}}{- 7.8}$

We now use a calculator.

$d = 3.927 m \mathmr{and} - 0.568$ m

A negative answer is clearly impossible so Alex will land $3.9$ meters away from the ramp. The graph of the $h \left(d\right)$ function confirms.

Hopefully this helps!

Mar 14, 2018

$19.7$ metres in height.

#### Explanation:

It's stated in this problem that Alex's path can be modelled by the quadratic function $h \left(d\right) = - 3.9 {d}^{2} + 13.1 d + 8.7$.

The maximum value of this function will be the maximum height reached by Alex.

In order to find the maximum value of $h \left(d\right)$, we need to complete the square:

$R i g h t a r r o w h \left(d\right) = - 3.9 {d}^{2} + 13.1 d + 8.7$

$R i g h t a r r o w h \left(d\right) = - 3.9 \left({d}^{2} + \frac{13.1}{- 3.9} d + \frac{8.7}{- 3.9}\right)$

$R i g h t a r r o w h \left(d\right) = - 3.9 \left({d}^{2} - \frac{13.1}{3.9} d - \frac{8.7}{3.9}\right)$

$R i g h t a r r o w h \left(d\right) = - 3.9 \left({d}^{2} - \frac{13.1}{3.9} d + {\left(\frac{\frac{13.1}{3.9}}{2}\right)}^{2} - \frac{8.7}{3.9} - {\left(\frac{\frac{13.1}{3.9}}{2}\right)}^{2}\right)$

$R i g h t a r r o w h \left(d\right) = - 3.9 \left({d}^{2} - \frac{13.1}{3.9} d + {\left(\frac{13.1}{7.8}\right)}^{2} - \frac{8.7}{3.9} - {\left(\frac{13.1}{7.8}\right)}^{2}\right)$

$R i g h t a r r o w h \left(d\right) = - 3.9 \left({\left(d - \frac{171.61}{60.84}\right)}^{2} - 5.0514464169\right)$

$R i g h t a r r o w h \left(d\right) = - 3.9 {\left(d - \frac{171.61}{60.84}\right)}^{2} + 19.70064103$

Now, the expression ${\left(d - \frac{171.61}{60.84}\right)}^{2}$ is always positive, as it is squared, i..e it is always greater than or equal to zero.

$R i g h t a r r o w {\left(d - \frac{171.61}{60.84}\right)}^{2} \ge q 0$

Multiplying this expression by $- 3.9$ (a negative number) reverses the inequality:

$R i g h t a r r o w - 3.9 {\left(d - \frac{171.61}{60.84}\right)}^{2} \le q 0$

Let's add $19.70064103$ to both sides of the inequality:

$R i g h t a r r o w - 3.9 {\left(d - \frac{171.61}{60.84}\right)}^{2} \le q 19.70064103$

This expression is now equivalent to our function $h \left(d\right)$ (in vertex form):

$\therefore h \left(d\right) \le q 19.70064103$

According to this inequality, the function $h \left(d\right)$ is less than or equal to $19.70064103$, i.e. has a maximum value of $19.70064103$.

Rounding this number to the nearest tenth gives $h \left(d\right) \le q 19.7$.

Therefore, Alex reaches no higher than $19.7$ metres.