Graphing Parabolas Practice

1. Graph the equation.

   $y=-2(x-1)^2-4$

   

2. Graph the function.

   $g(x)=2(x+4)(x)$

   
   

3. Graph the equation.

   $y=4x^2+8x+7$

   
   

4. Graph the function.

   $f(x)=2x^2-5$

   
   

Source: Khan Academy, https://www.khanacademy.org/math/algebra/x2f8bb11595b61c86:quadratic-functions-equations/x2f8bb11595b61c86:quadratic-forms-features/e/graphing_parabolas_2
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1. 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:

   
   

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2. 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:

   
   

   ---

3. 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:

   
   

   ---

4. 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: