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: