Graphing Quadratic Equations in Standard Form

1. \(15\) meters

The height of the stone at the time it is thrown is given by \(h(0)\).

\(\begin{aligned}

h(0) &=-5(0)^{2}+10(0)+15 \\

&=0+0+15 \\

&=15

\end{aligned}\)

In conclusion, the height of the stone at the time of throwing is \(15\) meters.

2.

The garden's area is modeled by a quadratic function, whose graph is a parabola.

The maximum area is reached at the vertex.

So in order to find when that happens, we need to find the vertex's \(w\)-coordinate.

The vertex's \(x\)-coordinate is the average of the two zeros, so let's find those first.

\(\begin{gathered}

A(x)=0 \\

-x^{2}+100 x=0 \\

x^{2}-100 x=0 \\

x(x-100)=0 \\

\swarrow \searrow\\

x=0 \text { or } x-100=0 \\

x=0 \text { or } x=100

\end{gathered}\)

Now let's take the zeros' average:

\(\frac{(0)+(100)}{2}=\frac{100}{2}=50\)

In conclusion, the maximum garden area occurs when the width is \(50\) meters.

3. \(8\) thousand dollars.

The company's profit is modeled by a quadratic function, whose graph is a parabola.

The maximum profit is reached at the vertex.

So in order to find the maximum profit, we need to find the vertex's \(y\)-coordinate.

We will start by finding the vertex's \(x\)-coordinate, and then plug that into \(P(x)\).

The vertex's \(x\)-coordinate is the average of the two zeros, so let's find those first.

\( \begin{gathered}

P(x) =0 \\

-2 x^{2}+16 x-24 =0 \\

x^{2}-8 x+12 =0 \\

(x-2)(x-6) =0\\

\swarrow \searrow \\

x-2=0 \text { or } x-6=0 \\

x=2 \text { or } x=6

\end{gathered} \)

Now let's take the zeros' average:

\(\frac{(2)+(6)}{2}=\frac{8}{2}=4\)

The vertex's \(x\)-coordinate is \(4\). Now let's find \(P(4)\):

\(\begin{aligned}

P(4) &=-2(4)^{2}+16(4)-24 \\

&=-32+64-24 \\

&=8

\end{aligned}\)

In conclusion, the maximum profit is \(8\) thousand dollars.

4. \(4\) seconds

The ball hits the ground when \(h(x)=0\).

\( \begin{gathered}

h(x)=0 \\

-2 x^{2}+4 x+16=0 \\

x^{2}-2 x-8=0 \\

(x+2)(x-4)=0 \\

\swarrow \searrow \\

x+2=0 \text { or } x-4=0 \\

x=-2 \text { or } x=4

\end{gathered} \)

We found that \(h(x)=0\) for \(x=-2\) or \(x=4\). Since \(x=-2\) doesn't make sense in our context, the only reasonable answer is \(x=4\).

In conclusion, the ball will hit the ground after \(4\) seconds.