# What is the equation of the parabola that has a vertex at  (-1, 16)  and passes through point  (3,20) ?

##### 2 Answers
Mar 5, 2018

$f \left(x\right) = \frac{1}{4} {\left(x + 1\right)}^{2} + 16$

#### Explanation:

The standard form of the equation of a parabola is:
$f \left(x\right) = a {\left(x - h\right)}^{2} + k$

From the question we know two things.

1. The parabola has a vertex at $\left(- 1 , 16\right)$
2. The parabola passes through the point $\left(3 , 20\right)$

With those two pieces of information, we can construct our equation for the parabola.

Let's start off with the basic equation:
$f \left(x\right) = a {\left(x - h\right)}^{2} + k$

Now we can substitute our vertex coordinates for $h$ and $k$

The $x$ value of your vertex is $h$ and the $y$ value of your vertex is $k$:

$f \left(x\right) = a {\left(x + 1\right)}^{2} + 16$

Note that putting $- 1$ in for $h$ makes it $\left(x - \left(- 1\right)\right)$ which is the same as $\left(x + 1\right)$

Now substitute the point the parabola passes through for $x$ and $y$ (or $f \left(x\right)$):

$20 = a {\left(x + 1\right)}^{2} + 16$

Looks good. Now we have to find $a$

Combine all like terms:

Add 3 + 1 inside the parentheses:
$20 = a {\left(4\right)}^{2} + 16$

Square 4:
$20 = 16 a + 16$

Factor out 16:
$20 = 16 \left(a + 1\right)$

Divide both sides by 16:
$\frac{20}{16} = a + 1$

Simplify $\frac{20}{16}$:
$\frac{5}{4} = a + 1$

Subtract 1 from both sides:
$\frac{5}{4} - 1 = a$

The LCD of 4 and 1 is 4 so $1 = \frac{4}{4}$:
$\frac{5}{4} - \frac{4}{4} = a$

Subtract:
$\frac{1}{4} = a$

Switch sides if you want:
$a = \frac{1}{4}$

Now that you've found $a$, you can plug it into the equation with the vertex coordinates:

$f \left(x\right) = \frac{1}{4} {\left(x + 1\right)}^{2} + 16$

And that's your equation.
Hope this helped.

Mar 5, 2018

$y = \frac{1}{4} {\left(x + 1\right)}^{2} + 16$

#### Explanation:

$\text{the equation of a parabola in "color(blue)"vertex form}$ is.

$\textcolor{red}{\overline{\underline{| \textcolor{w h i t e}{\textcolor{b l a c k}{y = a {\left(x - h\right)}^{2} + k}} \textcolor{w h i t e}{\frac{2}{2}} |}}}$

$\text{where "(h,k)" are the coordinates of the vertex and a}$
$\text{is a multiplier}$

$\text{here } \left(h , k\right) = \left(- 1 , 16\right)$

$\Rightarrow y = a {\left(x + 1\right)}^{2} + 16$

$\text{to find a substitute "(3,20)" into the equation}$

$20 = 16 a + 16 \Rightarrow a = \frac{1}{4}$

$\Rightarrow y = \frac{1}{4} {\left(x + 1\right)}^{2} + 16 \leftarrow \textcolor{red}{\text{in vertex form}}$