SOLUTION: Consider the general quadratic function f(x)=a(x-h)^2+k, where a, h, and k are real constants and (a) does not equal 0. Where does the graph of f(x) cross the y-axis? And if the ve
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Question 2386: Consider the general quadratic function f(x)=a(x-h)^2+k, where a, h, and k are real constants and (a) does not equal 0. Where does the graph of f(x) cross the y-axis? And if the vertex of the graph of f(x) is located at the point (2,3), how many x-intercepts are possible for the graph?
Answer by longjonsilver(2297) (Show Source): You can put this solution on YOUR website!
this has a +ve x^2 term therefore the graph is a u-shape.
a(x-h)^2+k=0
a(x-h)^2 = -k
(x-h)^2 = -(k/a)
x-h = +- sqrt(-(k/a))
x = h +- sqrt(-(k/a))
these are your 2 x-values...the roots.
if vertex is at (2,3) and we know it looks like a "U" then lowest point is at (2,3)...so it DOES NOT cross the x-axis..No x-intercepts.
jon
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