Graphing Quadratic Equations in Vertex Form

1. Graph the equation.

$y=4(x-2)^{2}-6$

2. Graph the function.

$g(x)=2(x-2)^{2}+2$

3. Graph the equation.

$y=5(x+1)^{2}-3$

4. Graph the function.

$f(x)=-3(x+1)^{2}+5$

1.

The strategy

The equation is in vertex form $y=a(x-h)^{2}+k$.

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

• In vertex form, the vertex coordinates are simply $(h, k)$.
  
• The other point can be a point next to the vertex $(\text { where } x=h \pm 1)$

Finding the vertex

The coordinates of the vertex of a parabola in the form $y=a(x-h)^{2}+k \text { are }(h, k)$

Note that $h$ is found when it is subtracted from $x$. For this reason, let's rewrite the given equation as follows:

$y=4(x-2)^{2}-6$

Therefore, the vertex is at $(2,6)$.

Finding another point

When the equation is given in vertex form, it's usually best to look for another point that is near the vertex. Since the vertex is at $x=2$, let's plug $x=1$ into the equation.

$\begin{aligned} y &=4(1-2)^{2}-6 \\ &=4(-1)^{2}-6 \\ &=4-6 \\ &=-2 \end{aligned}$

Therefore, another point on the parabola is $(1,-2)$.

The solution

The vertex of the parabola is at $(2,-6)$ and another point on the parabola is at $(1,-2)$.

Therefore, this is the parabola:

2.

The strategy

The equation is in vertex form $g(x)=a(x-h)^{2}+k$.

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

• In vertex form, the vertex coordinates are simply $(h, k)$.
• The other point can be a point next to the vertex $(\text { where } x=h \pm 1)$

Finding the vertex

The coordinates of the vertex of a parabola in the form $g(x)=a(x-h)^{2}+k \text { are }(h, k)$

Note that $h$ is found when it is subtracted from $x$. For this reason, let's rewrite the given equation as follows:

$g(x)=2(x-2)^{2}+2$

Therefore, the vertex is at $2,2)$.

Finding another point

When the equation is given in vertex form, it's usually best to look for another point that is near the vertex. Since the vertex is at $x=2$, let's plug $x=1$ into the equation.

$\begin{aligned} g(x) &=2(1-2)^{2}+2 \\ &=2(-1)^{2}+2 \\ &=2+2 \\ &=4 \end{aligned}$

Therefore, another point on the parabola is $(1,4)$.

The solution

The vertex of the parabola is at $(2,2)$ and another point on the parabola is at $(1,4)$.

Therefore, this is the parabola:

3.

The strategy

The equation is in vertex form $g(x)=a(x-h)^{2}+k$.

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

• In vertex form, the vertex coordinates are simply $(h, k)$.
• The other point can be a point next to the vertex $(\text { where } x=h \pm 1)$

Finding the vertex

The coordinates of the vertex of a parabola in the form $y=a(x-h)^{2}+k \text { are }(h, k)$

Note that $h$ is found when it is subtracted from $x$. For this reason, let's rewrite the given equation as follows:

$y=5(x-(-1))^{2}-3$

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

Finding another point

When the equation is given in vertex form, it's usually best to look for another point that is near the vertex. Since the vertex is at $x=-1$, let's plug $x=0$ into the equation.

$\begin{aligned} y &=5(0+1)^{2}-3 \\ &=5(1)^{2}-3 \\ &=5-3 \\ &=2 \end{aligned}$

Therefore, another point on the parabola is0 $(0,2)$.

The solution

The vertex of the parabola is at $(-1,-3)$ and another point on the parabola is at $(0,2)$.

Therefore, this is the parabola:

4.

The strategy

The equation is in vertex form $y=a(x-h)^{2}+k$.

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

• In vertex form, the vertex coordinates are simply $(h, k)$.
  
• The other point can be a point next to the vertex $(\text { where } x=h \pm 1)$

Finding the vertex

The coordinates of the vertex of a parabola in the form $g(x)=a(x-h)^{2}+k \text { are }(h, k)$

Note that $h$ is found when it is subtracted from $x$. For this reason, let's rewrite the given equation as follows:

$f(x)=-3(x-(-1))^{2}+5$

Therefore, the vertex is at $(-1,5)$.

Finding another point

When the equation is given in vertex form, it's usually best to look for another point that is near the vertex. Since the vertex is at $x=-1$, let's plug $x=0$ into the equation.

$\begin{aligned} f(x) &=-3(0+1)^{2}+5 \\ &=-3(1)^{2}+5 \\ &=-3+5 \\ &=2 \end{aligned}$

Therefore, another point on the parabola is $(0,2)$.

The solution

The vertex of the parabola is at $(-1,5)$ and another point on the parabola is at $(0,2)$.

Therefore, this is the parabola:

1. A hovercraft takes off from a platform.

Its height (in meters), $x$ seconds after takeoff, is modeled by:

$h(x)=-3(x-3)^{2}+108$

What is the height of the hovercraft at the time of takeoff?

2.A certain company's main source of income is selling socks.

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

$P(x)=-3(x-5)^{2}+12$

What sock price should the company set to earn a maximum profit?

3.The fish population in a certain part of the ocean (in thousands of fish) as a function of the water's temperature (in degrees Celsius) is modeled by:

$P(x)=-2(x-9)^{2}+200$

What is the maximum number of fish?

4. The power generated by an electrical circuit (in watts) as a function of its current $c$ (in amperes) is modeled by:

$P(c)=-20(c-3)^{2}+180$

Which currents will produce no power (i.e. 000 watts)?
Enter the lower current first.

Lower current: ______ amperes

Higher current: ______ amperes

1. $81$ meters

The height of the hovercraft at the time of takeoff is given by $h(0)$.

$\begin{aligned} h(0) &=-3(0-3)^{2}+108 \\ &=-3(9)+108 \\ &=81 \end{aligned}$

In conclusion, the height of the hovercraft at the time of takeoff is $81$ meters.

2. $5$ 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 when that happens, we need to find the vertex's $x$-coordinate.

The function $P(x)$ is given in vertex form.

The vertex of $-3(x-5)^{2}+12 \text { is at }(5,12)$.

In conclusion, the company will earn a maximum profit when the socks are priced at $5$ dollars.

3. $200$ thousand fish

The fish population is modeled by a quadratic function, whose graph is a parabola.

The maximum number of fish is reached at the vertex.

So in order to find the maximum number of fish, we need to find the vertex's $y$-coordinate.

The function $P(x)$ is given in vertex form.

The vertex of $-2(x-9)^{2}+200$ is at $(9,200)$.

In conclusion, the maximum fish population is $200$ thousand.

4. Lower current: $0$ amperes, Higher current: $6$ amperes

The circuit's power is $0$ when $P(c)=0$.

$\begin{gathered} P(c)=0 \\ -20(c-3)^{2}+180=0 \\ -20(c-3)^{2}=-180 \\ (c-3)^{2}=9 \\ \sqrt{(c-3)^{2}}=\sqrt{9} \\ c-3=\pm 3 \\ c=\pm 3+3 \\ c=6 \text { or } c=0 \end{gathered}$

In conclusion, these are the currents that will produce no power:

Lower current: $0$ amperes

Higher current: $6$ amperes