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

tinfoQ 2021-02-09 Answered

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

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Expert Answer

Khribechy
Answered 2021-02-10 Author has 100 answers

The intercepts are found by puttin gx=0(y) and y=0(x) into the parabola equation: y=(x+1)2+1.
When x=0, y=1+1=0, when y=0, (x+1)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.

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