# How do you find the vertex and intercepts for y = (1/8)(x – 5)^2 - 3?

Jul 22, 2018

$\text{see explanation}$

#### Explanation:

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

•color(white)(x)y=a(x-h)^2+k

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

$y = \frac{1}{8} {\left(x - 5\right)}^{2} - 3 \text{ is in this form}$

$\textcolor{m a \ge n t a}{\text{vertex }} = \left(5 , - 3\right)$

$\text{to find y-intercept let x = 0}$

$y = \frac{25}{8} - \frac{24}{8} = \frac{1}{8} \leftarrow \textcolor{red}{\text{y-intercept}}$

$\text{to find x-intercepts let y = 0}$

$\frac{1}{8} {\left(x - 5\right)}^{2} - 3 = 0$

$\frac{1}{8} {\left(x - 5\right)}^{2} = 3 \Rightarrow {\left(x - 5\right)}^{2} = 24$

$\textcolor{b l u e}{\text{take the square root of both sides}}$

$x - 5 = \pm \sqrt{24} = \pm 2 \sqrt{6}$

$\text{add 5 to both sides}$

$x = 5 \pm 2 \sqrt{6} \leftarrow \textcolor{red}{\text{exact solutions}}$

$x \approx 0.1 , x \approx 9.9 \text{ to 1 dec. place}$

Vertex : $\left(5 , - 3\right)$,

x-intercepts : $= 5 \setminus \pm 2 \setminus \sqrt{6}$ &

y-intercept: $= \frac{1}{8}$

#### Explanation:

Given equation:

$y = \left(\frac{1}{8}\right) {\left(x - 5\right)}^{2} - 3$

$\frac{1}{8} {\left(x - 5\right)}^{2} = y + 3$

${\left(x - 5\right)}^{2} = 8 \left(y + 3\right)$

The above equation is in form of vertical parabola: ${\left(x - {x}_{1}\right)}^{2} = 4 a \left(y - {y}_{1}\right)$ which has

Vertex $\left({x}_{1} , {y}_{1}\right) \setminus \equiv \left(5 , - 3\right)$

Setting $x = 0$ in the above equation of parabola:

${\left(0 - 5\right)}^{2} = 8 \left(y + 3\right)$

$y = \frac{1}{8}$

The given parabola intersects the y-axis at $\left(0 , \frac{1}{8}\right)$ &

y-intercept: $= \frac{1}{8}$

Similarly, setting $y = 0$ in above equation of parabola, we get

${\left(x - 5\right)}^{2} = 8 \left(0 + 3\right)$

$x - 5 = \setminus \pm \setminus \sqrt{24}$

$x = 5 \setminus \pm 2 \setminus \sqrt{6}$

The given parabola intersects the x-axis at two points

$\left(5 \setminus \pm 2 \setminus \sqrt{6} , 0\right)$

x-intercepts : $= 5 \setminus \pm 2 \setminus \sqrt{6}$

Jul 22, 2018

The vertex is $\left(5 , - 3\right)$.
The y-intercept is $\left(0 , \frac{1}{8}\right)$ or $\left(0 , 0.125\right)$.
The x-intercepts are $\left(5 + 2 \sqrt{6} , 0\right)$ and $\left(5 - 2 \sqrt{6} , 0\right)$.
The approximate x-intercepts are $\left(9.899 , 0\right)$ and $\left(0.101 , 0\right)$.

#### Explanation:

$y = \frac{1}{8} {\left(x - 5\right)}^{2} - 3$ is a quadratic equation in vertex form:

$y = a {\left(x - h\right)}^{2} + k$,

where:

$h = 5$ and $k = - 3$

The vertex is $\left(h , k\right)$, which is $\left(5 , - 3\right)$.

To find the y-intercept, substitute $0$ for $x$ and solve for $y$.

$y = \frac{1}{8} {\left(0 - 5\right)}^{2} - 3$

$y = \frac{1}{8} \left(25\right) - 3$

$y = \frac{25}{8} - 3$

Simplify $3$ to $\frac{24}{8}$.

$y = \frac{25}{8} - \frac{24}{8} = \frac{1}{8} = 0.125$

The y-intercept is $\left(0 , \frac{1}{8}\right)$ or $\left(0 , 0.125\right)$.

To find the x-intercepts, substitute $0$ for $y$ and solve.

$0 = \frac{1}{8} {\left(x - 5\right)}^{2} - 3$

Multiply both sides by $8$.

$0 = {\left(x - 5\right)}^{2} - 24$

Expand ${\left(x - 5\right)}^{2}$.

$0 = {x}^{2} - 10 x + 25 - 24$

$0 = {x}^{2} - 10 x + 1$,

where:

$a = 1$, $b = - 10$, $c = 1$

Use the quadratic formula.

$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

Plug in the known values and solve.

$x = \frac{- \left(- 10\right) \pm \sqrt{{\left(- 10\right)}^{2} - 4 \cdot 1 \cdot 1}}{2 \cdot 1}$

$x = \frac{10 \pm \sqrt{96}}{2}$

Prime factorize $96$.

$x = \frac{10 \pm \sqrt{2 \times 2 \times 2 \times 2 \times 2 \times 3}}{2}$

$x = \frac{10 \pm \sqrt{{2}^{2} \times {2}^{2} \times 2 \times 3}}{2}$

$x = \frac{10 \pm 2 \times 2 \sqrt{2 \times 3}}{2}$

$x = \frac{10 \pm 4 \sqrt{6}}{2}$

Factor out the common $2$.

$x = 5 \pm 2 \sqrt{6}$

$x = 5 + 2 \sqrt{6}$, $5 - 2 \sqrt{6}$

Approximate values for $x$:

$x \approx 9.899 , 0.101$

The x-intercepts are $\left(5 + 2 \sqrt{6} , 0\right)$ and $\left(5 - 2 \sqrt{6} , 0\right)$.

The approximate x-intercepts are $\left(9.899 , 0\right)$ and $\left(0.101 , 0\right)$.

graph{y=1/8(x-5)^2-3 [-4.29, 15.71, -4.64, 5.36]}