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: