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When solving quadratics by factoring, refer to the following formulas, as
needed.
acx
2 (ad bc)x bd = (ax b)(cx d) General trinomial
x
2 2xy y
2 = (x y)
2 Perfect square
x
2 – y
2 = (x y)(x – y) Difference of two squares
x
2 y
2 = (x yi)(x – yi) Sum of two squares
Note: Recall that i
2 = –1.
197. Solve x
2 – x – 6 = 0 by factoring.
198. Solve x
2 6x – 1 = – 5 by completing the square.
199. Solve 3x
2 – 5x 1 = 0 by using the quadratic formula.
For questions 200 to 202, solve using any preferred method.
200. x
2 – 3x 2 = 0
201. 9x
2 18x – 17=0
202. 6x
2 – 12x 7 = 0
For questions 203 to 207, solve the quadratic inequality. Write the solution set in
interval notation. Refer to the following guidelines, as needed.
To solve a quadratic inequality with a > 0, arrange terms so that only 0 is on the
right side of the inequality and apply the following:
If ax
2 bx c = 0 has two real roots, ax
2 bx c is negative between them,
positive to the left of the leftmost root, positive to the right of the rightmost
root, and zero only at the roots.
If ax
2 bx c = 0 has exactly one real root, ax
2 bx c is zero at that root
and positive elsewhere.
If ax
2 bx c = 0 has no real roots, ax
2 bx c is always positive.
Note: If you have a quadratic inequality in which a 0.
203. x
2 – x – 12 < 0
204. –x
2 x 12 0
206. x
2 – 10x 25 ≥ 0
207. x
2 – 5 ≥ 0
Average Rate of Change and Difference Quotients for Linear and
Quadratic Functions
For questions 208 to 213, find the average rate of change of f on the given
interval.
Note: The average rate of change of f as x goes from x1
to x2
is .
208. f(x) = –2x – 5, [–3,3]
209. f(x) = x 3, [–4,8]
210. f(x) = 10, [–5,5]
211. f(x) = kx, [–3,5]
212. f(x) = x
2 8x – 5, [–8,–4]
213. f(x) = x
2 8x – 5, [–4,8]
214. Prove that the average rate of change of a linear function f defined by f(x) =
mx b is constant over any interval [c, d] and equals the slope m of the
graph of f.
215. Prove that the average rate of change of a quadratic function f defined by
f(x) = ax
2 bx c over an interval [m, n] equals a(n m) b.
For questions 216 to 219, find and simplify the difference quotient for the
function.
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Note: The difference quotient is the expression , where h ≠ 0. This
expression is often used in calculus to find a general expression for the average
rate of change of a function. It is the average rate of change of f as x goes from x
to x h.
216. f(x) = 2x – 5
217. g(x) = 2x 50
218. f(x) = x
2 3x
219. f(x) = ax
2 bx c
220. Prove that the difference quotient of a linear function f defined by f(x) = mx
b equals the slope m of the graph of f.
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