Graphing Quadratic Equations in Standard Form

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: