Graphing Parabolas Practice

This equation is in vertex form.

$y={a}(x-h)^2+k$

This form reveals the vertex, ‍$(h,k)$, which in our case is ‍$(1,-4)$.

It also reveals whether the parabola opens up or down. Since ‍$a=-2$, the parabola opens downward.

This is enough to start sketching the graph.

To finish our graph, we need to find another point on the curve.

Let's plug $x=0$ into the equation.

$\begin{aligned}y&=-2(x-1)^2-4\\\\&=-2(0-1)^2-4\\\\&=-2(1)-4\\\\&=-2-4\\\\&=-6\end{aligned}$

Therefore, another point on the parabola is $(0,-6)$.

The answer:

The strategy

The function is in the factored form $g(x)=a(x-x_1)(x-x_2)$. In this form, the ‍$x$-intercepts of the parabola are at ‍$(x_1,0)$ and $(x_2,0)$.

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

• The vertex can be found by using the fact that it lies on the vertical line exactly between the $x$-intercepts.

• The other point can be one of the $x$-intercepts.

Finding the $x$-intercepts

The $x$-intercepts of a parabola in the form $g(x)=a(x-{x_1})(x-{x_2})$ are at ‍$x={x_1}$ and $x={x_2}$.

Note that ${x_1}$ and ‍${x_2}$ are found when they are subtracted from $x$. For this reason, let's rewrite the given equation as follows:

$g(x)=2(x-({-4}))(x-({0}))$

Therefore, the $x$-intercepts are at ‍$x={-4}$ and $x={0}$.

Finding the vertex

The vertex lies on the parabola's axis of symmetry, which is exactly between the two $x$-intercepts. This means that its ‍$x$-coordinate is the average of the two $x$-intercepts.

$\begin{aligned}\dfrac{{x_1}+{x_2}}{2}&=\dfrac{{-4}+{0}}{2}\\\\&=\dfrac{-4}{2}\\\\&=-2\end{aligned}$

We can now plug $x=-2$ into the function to find the $x$-coordinate of the vertex:

$\begin{aligned}g(x)&=2(-2+4)(-2)\\\\&=2\cdot (2)(-2)\\\\&=-8\end{aligned}$

In conclusion, the vertex is at $(-2, -8)$.

The solution

The vertex of the parabola is at $(-2, -8)$ and other points on the parabola are ‍$(-4, 0)$ and $(0, 0)$.

Therefore, this is the parabola:

The strategy

The equation is in the standard form $y=ax^2+bx+c$.

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

• The vertex can be found by using the formula for the $x$-coordinate, $\ -\dfrac{b}{2a}$.

• The other point can be one of the $y$-intercepts, which in standard form is simply $(0,c)$.

Finding the vertex

The ‍$x$-coordinate of the vertex of a parabola in the form ‍$y=ax^2+bx+c$ is $-\dfrac{b}{2a}$.

Our equation is ‍$y={4}x^2+{8}x+{7}$, so this is the ‍$x$-coordinate of its vertex:

$\begin{aligned}-\dfrac{{8}}{2\cdot{4}}&=-\dfrac{8}{8}\\\\&=-1\end{aligned}$

We can now plug $x=-1$ into the equation to find the $y$-coordinate of the vertex:

$\begin{aligned}y&=4(-1)^2+8(-1)+7\\\\&=4\cdot 1-8+7\\\\&=3\end{aligned}$

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

Finding the $y$-intercept

The $y$-intercept of a parabola in the form $y=ax^2+bx+c$ are at ‍$x={x_1}$ is $(0, c)$.

Our equation is ‍$y={4}x^2+{8}x+{7}$, so its ‍‍$y$-intercept is ‍‍$(0, {7})$.

The solution

The vertex of the parabola is at $(-1, 3)$ and the $y$-intercept is at $(0, 7)$.

Therefore, this is the parabola:

The strategy

This equation is in vertex form $f(x)=a(x-h)^2+k$.

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 (‍where $x=h\pm 1$).

Finding the vertex

The ‍coordinates of the vertex of a parabola in the form ‍$f(x)=a(x-h)^2+k$ 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:

$f(x)=2(x-{0})^2{-5}$

Therefore, the vertex is at $({0},{-5})$.

Finding the $y$-intercept

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 ‍$x=0$, let's plug ‍$x=1$ into the equation.

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

Therefore, another point on the parabola is $(1, -3)$.

The solution

The vertex of the parabola is at $(0, -5)$ and another point on the parabola is at $(1, -3)$.

Therefore, this is the parabola: