# What is the process to solve these:Vertex, y-int., x-int, graph= -(x+1)^2+1

What is the process to solve these:
Vertex, $y-\int$., $x-\int$, graph $=-\left(x+1{\right)}^{2}+1$

You can still ask an expert for help

• Questions are typically answered in as fast as 30 minutes

Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it

Khribechy

The intercepts are found by puttin $gx=0\left(y-\int \right)$ and $y=0\left(x-\int \right)$ into the parabola equation: $y=-{\left(x+1\right)}^{2}+1.$
When $x=0$, $y=-1+1=0$, when $y=0$, ${\left(x+1\right)}^{2}=1$, so $x+1=+-1$, and $x=-2$ and 0.
That gives us 0 as y-int and -2 and 0 as x-ints. The graph is an inverted U shape where the arms of the U intersect the x-axis at -2 and 0, so the U lies to the left of the y axis.
The x value for the vertex is midway between the x intercepts at $x=-1$ and $y=1\left(-1,1\right)$, and this can also be seen in the term $\left(x+1\right)$ as the displacement from the origin.