You can put this solution on YOUR website! find the range and domain and vertex and intercepts of a graph whose parabola is x^2+4x-5
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Standard form of parabola: y=(x-h)^2+k, with (h,k) being the (x,y) coordinates of the vertex.
y=x^2+4x-5
completing the square
y=(x^2+4x+4)-5-4
y=(x+2)^2-9
Parabola opens upwards with axis of symmetry at x=-2
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Vertex: (-2,-9)
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y-intercept
x=0
y=-5
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x-intercept
y=0=x^2+4x-5
(x+5)(x-1)=0
x=-5
or x=1
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Range:[-9,∞)
Domain:(-∞,∞)
See graph below for a visual check on the answers above.
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y=x^2+4x-5
(