Graphing Quadratic Equations in Standard Form

1. Graph the equation.

$y=2 x^{2}-8 x+3$

2. Graph the function.

$h(x)=x^{2}+2 x$

3. Graph the equation.

$y=-\frac{1}{8} x^{2}+x-4$

4. Graph the function.

$h(x)=-\frac{1}{3} x^{2}+2 x-4$

1.

The strategy

The equation is in the standard form $y=a x^{2}+b x+c$.

To graph the parabola, we need its vertex and another point on the parabola.

• The vertex can be found using the formula for the $x$-coordinate, $-\frac{b}{2 a}$
• The other point can be the $y$-intercept, which in standard form is simply $(0, c)$.

Finding the vertex

The $x$-coordinate of the vertex of a parabola in the form $a x^{2}+b x+c \text { is }-\frac{b}{2 a}$.

Is it possible to find the vertex without this formula?

You can always bring the equation to vertex form by completing the square.

In the vertex form $y=a(x-h)^{2}+k$, the vertex is at $(h,k)$

Our equation is $y=2 x^{2}-8 x+3$, so this is the $x$-coordinate of its vertex:

$\begin{aligned} -\frac{-8}{2 \cdot 2} &=\frac{8}{4} \\ &=2 \end{aligned}$

We can now plug $x=2$ into the equation to find the $y$-coordinate of the vertex:

$\begin{aligned} y &=2(2)^{2}-8(2)+3 \\ &=2 \cdot 4-16+3 \\ &=-5 \end{aligned}$

In conclusion, the vertex is at $(2,-5)$.

Finding the $y$-intercept

The $y$-intercept of a parabola in the form $a x^{2}+b x+c \text { is }(0, c)$.

Our equation is $y=2 x^{2}-8 x+3$, so it's $y$-intercept is $(0, 3)$.

The solution

The vertex of the parabola is at $(2,-5)$ right parenthesis and the $y$-intercept is at $(0,3)$.

Therefore, this is the parabola:

2.

The strategy

The equation is in the standard form $y=a x^{2}+b x+c$.

To graph the parabola, we need its vertex and another point on the parabola.

• The vertex can be found using the formula for the $x$-coordinate, $-\frac{b}{2 a}$
• The other point can be the $y$-intercept, which in standard form is simply $(0, c)$.

Finding the vertex

The $x$-coordinate of the vertex of a parabola in the form $a x^{2}+b x+c \text { is }-\frac{b}{2 a}$.

Is it possible to find the vertex without this formula?

You can always bring the equation to vertex form by completing the square.

In the vertex form $y=a(x-h)^{2}+k$, the vertex is at $(h,k)$

Our function is $h(x)=1 x^{2}+2 x+0$, so this is the $x$-coordinate of the vertex.

$\begin{aligned} -\frac{2}{2 \cdot 1} &=-\frac{2}{2} \\ &=-1 \end{aligned}$

We can now plug $x=-1$ into the equation to find the $y$-coordinate of the vertex:

$\begin{aligned} y &=1(-1)^{2}+2(-1) \\ &=1 \cdot 1+-2 \\ &=-1 \end{aligned}$

In conclusion, the vertex is at $(-1,-1)$.

Finding the $y$-intercept

The $y$-intercept of a parabola in the form $a x^{2}+b x+c \text { is }(0, c)$.

Our function is $h(x)=1 x^{2}+2 x+0$, so it's $y$-intercept is $(0,0)$.

The solution

The vertex of the parabola is at $(-1,-1)$ and the $y$-intercept is at $(0,0)$.

Therefore, this is the parabola:

3.

The strategy

The equation is in the standard form $y=a x^{2}+b x+c$.

To graph the parabola, we need its vertex and another point on the parabola.

• The vertex can be found using the formula for the $x$-coordinate, $-\frac{b}{2 a}$
• The other point can be the $y$-intercept, which in standard form is simply $(0, c)$.

Finding the vertex

The $x$-coordinate of the vertex of a parabola in the form $a x^{2}+b x+c \text { is }-\frac{b}{2 a}$.

Is it possible to find the vertex without this formula?

You can always bring the equation to vertex form by completing the square.

In the vertex form $y=a(x-h)^{2}+k$, the vertex is at $(h,k)$

Our equation is $y=-\frac{1}{8} x^{2}+1 x-4$, so this is the $x$-coordinate of its vertex:

$\begin{aligned} -\frac{1}{2 \cdot\left(-\frac{1}{8}\right)} &=\frac{1}{\frac{1}{4}} \\ &=\frac{1 \cdot 4}{1} \\ &=4 \end{aligned}$

We can now plug $x=4$ into the equation to find the $y$-coordinate of the vertex:

$\begin{aligned} y &=-\frac{1}{8}(4)^{2}+1(4)-4 \\ &=-\frac{1}{8} \cdot 16+4-4 \\ &=-2 \end{aligned}$

In conclusion, the vertex is at $(4,-2)$.

Finding the $y$-intercept

The $y$-intercept of a parabola in the form $a x^{2}+b x+c$ is $(0, c)$.

Our equation is $y=-\frac{1}{8} x^{2}+1 x-4$, so its $y$-intercept is $(0, -4)$.

The solution

The vertex of the parabola is at $(4, -2)$ and the $y$-intercept is at $(0,-4)$.

Therefore, this is the parabola:

4.

The strategy

The equation is in the standard form $y=a x^{2}+b x+c$.

To graph the parabola, we need its vertex and another point on the parabola.

• The vertex can be found using the formula for the xxx-coordinate, $-\frac{b}{2 a}$
• The other point can be the $y$-intercept, which in standard form is simply $(0, c)$.

Finding the vertex

The $x$-coordinate of the vertex of a parabola in the form $a x^{2}+b x+c \text { is }-\frac{b}{2 a}$.

Is it possible to find the vertex without this formula?

You can always bring the equation to vertex form by completing the square.

In the vertex form $y=a(x-h)^{2}+k$, the vertex is at $(h,k)$

$\begin{aligned} -\frac{2}{2 \cdot\left(-\frac{1}{3}\right)} &=\frac{2}{2} \\ &=\frac{2 \cdot 3}{2} \\ &=3 \end{aligned}$

We can now plug $x=3$ into the equation to find the $y$-coordinate of the vertex:

$\begin{aligned} y &=-\frac{1}{3}(3)^{2}+2(3)-4 \\ &=-\frac{1}{3} \cdot 9+6-4 \\ &=-1 \end{aligned}$

In conclusion, the vertex is at $(3,-1)$.

Finding the $y$-intercept

The $y$-intercept of a parabola in the form $a x^{2}+b x+c$ is $(0, c)$.

Our function is $h(x)=-\frac{1}{3} x^{2}+2 x-4$, so its $y$-intercept is $(0, -4)$

The solution

The vertex of the parabola is at $(3,-1)$ and the $y$-intercept is at $(0,-4)$.

Therefore, this is the parabola:

1. Alain throws a stone off a bridge into a river below.

The stone's height (in meters above the water), $x$ seconds after Alain threw it, is modeled by:

$h(x)=-5 x^{2}+10 x+15$

What is the height of the stone at the time it is thrown?

2. Marquise has $200$ meters of fencing to build a rectangular garden.

The garden's area (in square meters) as a function of the garden's width $x$ (in meters) is modeled by:

$A(x)=-x^{2}+100 x$

What side width will produce the maximum garden area?

3.A certain company's main source of income is selling cloth bracelets.

The company's annual profit (in thousands of dollars) as a function of the price of a bracelet (in dollars) is modeled by:

$P(x)=-2 x^{2}+16 x-24$

What is the maximum profit that the company can earn?

4.Antoine stands on a balcony and throws a ball to his dog, who is at ground level.

The ball's height (in meters above the ground), $x$ seconds after Antoine threw it, is modeled by:

$h(x)=-2 x^{2}+4 x+16$

How many seconds after being thrown will the ball hit the ground?

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.