Question

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

Answer 1

The intercepts are found by puttin \(gx=0 (y-\int)\) and \(y=0 (x-\int)\) into the parabola equation: \(\displaystyle{y}=-{\left({x}+{1}\right)}^{{2}}+{1}.\)
When \(x=0\), \(y=-1+1=0\), when \(y=0\), \(\displaystyle{\left({x}+{1}\right)}^{{2}}={1}\), 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 \(y=1 (-1,1)\), and this can also be seen in the term \((x+1)\) as the displacement from the origin.