Consider the problem of finding the line of symmetry and vertex of the quadratic equation f(x) =x^2-8x+15 what is the error in the solution below?
x^2-8x+15=0 x=-8/2 =-8/2=-4 line of symmetry
8^2-8x+15=0 (-4)^2-8(-4)+15=0
16+32+15=0 y=63
(-4,63) vertex
a. the solution is correct.
b. the line of symmetry should have been 4 instead of –4.
c. the vertex is incorrect; it should have been {–4, 53}.
d. -4^2 should have been squared as –16 instead of 16.

practice. practice. practice.

answer: b. f(x) = (x + 2)^3 (x^2 - 7x + 3)^4

step-by-step explanation:

polynomials always have as many roots as their order. this is a 4th order polynomial- hence 4 roots!

complex roots of polynomials always come in pairs. the quadratic equation has the "+/-" in that will alwaysgenerate m+ni amd m-ni, so 7-2i must be a root of this polynomial equation.

the order of the numerator and denominator are both 2. so the horizontal asymptote is y=a2/b2 =7

verticle asymptotes are the forbidden x-values. f(x) becomes undefined at x=-3 because this value will cause a zero denominator.

a graph of r(x) will have vertical asymptotes at ±1 because these values will make the denominator zero. so the domain of the function cannot include ±1.

you 're wlecome! : )

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