Graphing Quadratic Equations in Vertex Form

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: