# How do you find the x and y intercepts for y = -2(x-1)^2 + 3?

Apr 3, 2018

$y$ intercept is 1
$x$ intercepts are 2.22 and -0.22

#### Explanation:

The $y$ intercept is when $x = 0$
$\implies$ $y = - 2 {\left(0 - 1\right)}^{2} + 3$

$\implies$ $y = - 2$$\times$ 1 +3

$\implies$ $y = - 2 + 3$

$\implies$ $y = 1$

The $x$ intercept is when $y = 0$
$\implies$ $0 = - 2 {\left(x - 1\right)}^{2} + 3$

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

$\implies$ $0 = - 2 {x}^{2} + 4 x - 2 + 3$

$\implies$ $2 {x}^{2} - 4 x - 1 = 0$

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

$x = \frac{4 \setminus \pm \sqrt{16 + 8}}{4}$

$x = \frac{4 + \sqrt{24}}{4}$ or $x = \frac{4 - \sqrt{24}}{4}$
$x = 2.22474487$ or $x = - 0.22474487$

Apr 3, 2018

${p}_{{x}_{1}} \left(1 + \sqrt{\frac{3}{2}} | 0\right) \mathmr{and} {p}_{{x}_{2}} \left(1 - \sqrt{\frac{3}{2}} | 0\right)$
${p}_{y} \left(0 | 1\right)$

#### Explanation:

Lets start with the interception of the y-axis. At the point where the graph intercepts this axis, $x = 0$. Furthermore, it is the only point where $x = 0$.

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

It works the same way for the x-axis. The only difference is that $y = 0$ instead of $x = 0$.

$0 = - 2 {\left(x - 1\right)}^{2} + 3 | - 3$
-3 = -2(x-1)^2 |:(-2)#
$\frac{3}{2} = {\left(x - 1\right)}^{2} | \sqrt{}$
$\pm \sqrt{\frac{3}{2}} = x - 1 | + 1$
$1 \pm \sqrt{\frac{3}{2}} = x$
${p}_{1} \left(1 + \sqrt{\frac{3}{2}} | 0\right) \mathmr{and} {p}_{2} \left(1 - \sqrt{\frac{3}{2}} | 0\right)$

graph{y=-2(x-1)^2+3 [-2, 4, -5, 5]}