For what real numbers X is the value of x^2-6x+9 negative?
1 answer:
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Answer:
(4, -25)
Step-by-step explanation:
One way to answer this is to put the equation into vertex form.
f(x)= x^2 -8x -9 . . . . . given
Add and subtract the square of half the x-coefficient:
f(x) = x^2 -8x +(-8/2)^2 -9 -(-8/2)^2
f(x) = x^2 -8x +16 -25
f(x) = (x -4)^2 -25
Comparing this to the vertex form of a quadratic:
f(x) = a(x -h)^2 +k
we find that (h, k) = (4, -25). This is the vertex.
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<em>Alternate solution</em>
The line of symmetry for ...
f(x) = ax^2 +bx +c
is given by x = -b/(2a). For your given quadratic that line is ...
x = -(-8)/(2(1)) = 4
Evaluating f(4) gives the y-coordinate:
f(4) = 4^2 -8·4 -9 = -25
The vertex is (x, y) = (4, -25).
