How do you graph y= (x+1)^2 -10 ?

Aug 6, 2015

Three points make a curve. You can solve for the vertex (minimum) and the (presumably) two x-intercepts, and sketch whatever connects all three points.

Normally you might do this:
$x = - \frac{b}{2 a}$

But this form is easier. You basically have an ${x}^{2}$ curve.

This one is shifted left 1 unit and down 10 units from $\left(0 , 0\right)$, because $x + 1$ in parentheses indicates a shift opposite to the sign ($+$ is left, $-$ is right), and the $10$ outside of the parentheses corresponds to $+$ as up and $-$ as down.

So instead of the vertex at $\left(0 , 0\right)$, it is at $\textcolor{b l u e}{\left(\left(- 1 , - 10\right)\right)}$. The x-intercepts, you get from solving the equation set to $y = 0$:

$0 = {\left(x + 1\right)}^{2} - 10$
$\pm \sqrt{10} = x + 1$

$\textcolor{b l u e}{{x}_{\text{right}}} = \sqrt{10} - 1 \textcolor{b l u e}{\approx 2.162}$
$\textcolor{b l u e}{{x}_{\text{left}}} = - \sqrt{10} - 1 \textcolor{b l u e}{\approx - 4.162}$

Finally, if you want to, you get the y-intercept by setting $x = 0$:
$y = {\left(0 + 1\right)}^{2} - 10 = - 9$, and it is $\textcolor{b l u e}{\left(\left(0 , - 9\right)\right)}$.

You can see that here by clicking on the intercepts and the vertex:

graph{(x+1)^2 - 10 [-5, 5, -15, 5]}