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# 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= -(x+1)^2+1

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Linear equations and graphs asked 2021-02-09

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

## Answers (1) 2021-02-10

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.

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