# How do you find the vertex and the intercepts for  f(x)=-4(x+1)^2+1?

The vertex is $\left(- 1 , 1\right)$ and the intercepts for the $x$ axis are $\left(- 1.5 , 0\right)$ and $\left(- 0.5 , 0\right)$ and the intercepts for the $y$ axis are $\left(0 , - 3\right)$.
Vertex? If you are using this structure of quadratic equations ($f \left(x\right) = a {\left(x - h\right)}^{2} + k$ ) which you are, then the vertex coordinate is $\left(h , k\right)$, and in your quadratic equation, that would be $\left(- 1 , 1\right)$.
To find the intercepts of $x$ and $y$ axis, you simply plug in $f \left(x\right) = 0$ into your equation, solve, find two possible answers for $x$, slap a zero, comma, and parentheses next to them and get the coordinates $\left(- 1.5 , 0\right)$ and $\left(- 0.5 , 0\right)$. Then, you plug in $x = 0$ and do the same for the y axis, in which you will get one answer that is (0, -3).