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Create your own quadratic equation that has an axis of symmetry of x = −1 and whose graph opens down. Then find the vertex, domain, range, and x-intercepts. Show all work to receive full credit.
2 answers
your parabola would be
y = a(x+1)^2 + c
where a would be a negative number.
So choose any values for a and c
e.g. one possible answer:
y = -2(x+1)^2 + 5
vertex: (-1,5)
domain: any real value of x
range : y ≤ 5
x-intercept: let y=0
-2(x+1)^2 +5=0
(x+1)^2 = 5/2 or 10/4
x+1 = ± √10/2
x = -1 ± √10/2
notice my little trick of making my denominator a perfect square, so it would come out "nice" after taking the square root ?
y = a(x+1)^2 + c
where a would be a negative number.
So choose any values for a and c
e.g. one possible answer:
y = -2(x+1)^2 + 5
vertex: (-1,5)
domain: any real value of x
range : y ≤ 5
x-intercept: let y=0
-2(x+1)^2 +5=0
(x+1)^2 = 5/2 or 10/4
x+1 = ± √10/2
x = -1 ± √10/2
notice my little trick of making my denominator a perfect square, so it would come out "nice" after taking the square root ?
1 answer