# What are the important points needed to graph y= -x^2+2x+4?

Jun 7, 2018

$x$-intercepts at $\left(1 - \sqrt{5} , 0\right)$ and $\left(1 + \sqrt{5} , 0\right)$, $y$-intercept at $\left(0 , 4\right)$ and a turning point at $\left(1 , 5\right)$.

#### Explanation:

So we have $y = - {x}^{2} + 2 x + 4$, and usually the sorts of 'important' points that are standard for including on sketches of quadratics are axis intercepts and the turning points.

To find the $x$-intercept, simply let $y = 0$, then:
$- {x}^{2} + 2 x + 4 = 0$

Then we complete the square (this will also help with finding the turning point).

${x}^{2} - 2 x + 1$ is the perfect square, then we subtract one again to maintain the equality:
$- \left({x}^{2} - 2 x + 1\right) + 1 + 4 = 0$
$\therefore - {\left(x - 1\right)}^{2} + 5 = 0$

This is the 'turning-point' form of the quadratic, so you can read your stationary point right off: $\left(1 , 5\right)$ (alternatively you could differentiate and solve $y ' = 0$).

Now just transpose the equation:
${\left(x - 1\right)}^{2} = 5$
$\therefore x - 1 = \pm \sqrt{5}$
$\therefore x = 1 \pm \sqrt{5}$

The $y$-intercept is easy, When $x = 0$, $y = 4$.

And there you have it!