# How do you graph the parabola y = -2(x -1)^2 + 3 using vertex, intercepts and additional points?

Aug 3, 2018

Refer to the explanation.

#### Explanation:

Graph:

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

This is a quadratic equation in vertex form:

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

where:

$a = - 2$, $h = 1$, $k = 3$

The vertex is the point $\left(h , k\right)$, which is $\left(1 , 3\right)$. Plot this point.

The y-intercept is the value of $y$ when $x = 0$. Substitute $0$ for $x$ and solve for $y$.

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

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

$y = - 2 \left(1\right) + 3$

$y = - 2 + 3$

$y = 1$

The y-intercept is $\left(0 , 1\right)$. Plot this point.

To find the x-intercepts, convert the equation from vertex form to standard form.

Expand ${\left(x - 1\right)}^{2}$ to ${x}^{2} - 2 x + 1$.

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

$y = - 2 {x}^{2} + 4 x - 2 + 3$

$y = - 2 {x}^{2} + 4 x + 1$

The x-intercepts are the values of $x$ when $y = 0$. Substitute $0$ for $y$ and solve for $x$.

$0 = - 2 {x}^{2} + 4 x + 1$,

where:

$a = - 2$, $b = 4$, $c = 1$

Switch sides.

$- 2 {x}^{2} + 4 x + 1 = 0$

Solve for the x-intercepts using the quadratic formula.

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

Plug in the known values.

$x = \frac{- 4 \pm \sqrt{{4}^{2} - 4 \cdot - 2 \cdot 1}}{2 \cdot - 2}$

Simplify.

$x = \frac{- 4 \pm \sqrt{24}}{- 4}$

Prime factorize $24$.

$x = \frac{- 4 \pm \sqrt{2 \times 2 \times 2 \times 3}}{- 4}$

Simplify $2 \times 2$ to ${2}^{2}$.

$x = \frac{- 4 \pm \sqrt{{2}^{2} \times 2 \times 3}}{- 4}$

Apply rule $\sqrt{{a}^{2}} = a$

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

$x = \frac{- 4 + 2 \sqrt{6}}{- 4}$, $\frac{- 4 - 2 \sqrt{6}}{- 4}$

Simplify.

$x = 1 - \frac{2 \sqrt{6}}{2}$, $1 + \frac{2 \sqrt{6}}{2}$

The x-intercepts are $\left(1 - \frac{\sqrt{6}}{2} , 0\right)$ and $\left(1 + \frac{\sqrt{6}}{2} , 0\right)$.

The approximate x-intercepts are $\left(- 0.225 , 0\right)$ and $\left(2.225 , 0\right)$.

Plot the x-intercepts.

Choose values for $x$ and solve for $y$.

$x = 2$

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

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

$y = - 2 \left(1\right) + 3$

$y = - 2 + 3$

$y = 1$

The additional point is $\left(2 , 1\right)$. Plot this point.

To find additional points, choose values for $x$ and solve for $y$.

Plot the points and sketch a parabola through the points. Do not connect the dots.

graph{y=-2(x-1)^2+3 [-10, 10, -5, 5]}