Pre-Calc Exam 1 Flashcards
Find y-intercept(s) and x-intercept(s) of y = 2(x^2) - x - 1
y-intercept: (0, -1) x-intercepts: (-1/2, 0) and (1, 0)
Find y-intercept(s) and x-intercept(s) of y = 3(x^2) - 12
y = 3(x^2) +0x - 12 y-intercept: (0, -12) x-intercepts: (2, 0) and (-2, 0)
Find y-intercept(s) and x-intercept(s) of y = x + 3
y-intercept: (0, 3) x-intercept: (-3, 0)
Find y-intercept(s) and x-intercept(s) of y = (x^2) + 6x + 9
y-intercept: (0, 9) x-intercept: (-3, 0)
Find y-intercept(s) and x-intercept(s) of y = 3(x^2) + 18x + 27
y-intercept: (0, 27) x-intercept: (-3, 0)
Find y-intercept(s) and x-intercept(s) of y = x^2 + 4x + 5
y-intercept: (0, 5) x-intercept: 0 = x^2 + 4x + 5 x = [-b +/- sqrt(b^2 -4ac)]/2a x = [-2 +/- sqrt(2^2 -4*1*5)]/(2*1) = [-2 +/- sqrt(4-20)]/2 therefore, x-int = undefined
Find the inverse function of f(x) = 2x + 3
f(x) = y = 2x+3 y -3 = 2x (y-3)/2 = x [f(x)-3]/2 = f’(x)
Find the inverse function of f(x) = x + 1
f(x) = y = x + 1 y = x + 1 y-1 = x f(x) - 1 = f’(x)
Find the inverse function of f(x) = 2(x^2) - x -1
f(x) = y = 2(x^2) - x -1 y = 2(x^2) - x - 1 y + 1 = 2(x^2) - x (y + 1)/2 = (x^2) - (x/2) + (1/4) (y+1)/2 + 1/4= (x+ 1/2)^2 sqrt[(y+1)/2 + 1/4] = sqrt[(x+ 1/2)^2] sqrt[(y+1)/2 + 1/4] = x+ 1/2 sqrt[(y+1)/2 + 1/4] - 1/2 = x sqrt{[f(x)+1]/2 + 1/4} - 1/2 = f’(x)
Find the inverse function of f(x) = x^2 + 6x + 9
f(x) = y = x^2 + 6x + 9 y = x^2 + 6x + 9 y = (x+3)^2 sqrt(y) = sqrt[(x+3)^2] sqrt(y) = x + 3 sqrt(y) - 3 = x sqrt[f(x)] - 3 = f’(x)
Find the inverse function of f(x) = x^2 -12x + 36
f(x) = y = x^2 -12x + 36 y = x^2 -12x + 36 y = (x-6)^2 sqrt(y) = sqrt[(x-6)^2] sqrt(y) = (x-6) sqrt(y) + 6 = x sqrt[f(x)] + 6 = f’(x)
Find the inverse function of f(x) = 3(x^2) + 5x + 25
f(x) = y = 3(x^2) + 5x + 25 y = 3(x^2) + 5x + 25 y - 25 = 3(x^2) + 5x (y - 25)/3 = [3(x^2) + 5x]/3 (y - 25)/3 = (x^2) + 5x/3 (y - 25)/3 + [(5/3)/2]^2 = (x^2) + 5x/3 + [(5/3)/2]^2 (y - 25)/3 + (5/6)^2 = (x^2) + 5x/3 + (5/6)^2 (y - 25)/3 + 25/36 = (x^2) + 5x/3 + 25/36 (y - 25)/3 + 25/36 = (x + 5/6)^2 sqrt[(y - 25)/3 + 25/36] = sqrt[(x + 5/6)^2] sqrt[(y - 25)/3 + 25/36] = x + 5/6 sqrt[(y - 25)/3 + 25/36] -5/6 = x sqrt{[f(x) - 25]/3 + 25/36} -5/6 = f’(x)
Determine whether the statement is true or false. If false, describe how the statement might be changed to make it true. 1. The solution to the equation 3x − 2 = 4 is x = 7/2
False because the solution to the equation 3x − 2 = 4 is x = 2.
Determine whether the statement is true or false. If false, describe how the statement might be changed to make it true. 2. The solutions to the equation x^2 − 3x + 2 = 0 are x = 2 and x = −1.
(x-1)(x-2)=0 x=1, 2 Therefore, false. The solutions to the equation x^2 − 3x + 2 = 0 are x=1 and 2.
Determine whether the statement is true or false. If false, describe how the statement might be changed to make it true. 3. The zeros of f (x) = x^2 + 2x − 4 are irrational numbers.
f(x) = (x^2 +2x) -4 f(x) = (x^2 + 2x +1) - 4 + 1 f(x) = (x-1)^2 -3 (x-1)^2 - 3 = 0 (x-1)^2 = 3 sqrt[(x-1)^2] = sqrt(3) x-1 = sqrt(3) x = sqrt(3)+1 Therefore, true.
Determine whether the statement is true or false. If false, describe how the statement might be changed to make it true. 4. The zeros of f(x) = 2(x^3) − x^2 − x are rational numbers.
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Determine whether the statement is true or false. If false, describe how the statement might be changed to make it true. 5. The solution set of the inequality 2x + 1
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Determine whether the statement is true or false. If false, describe how the statement might be changed to make it true. 6. The solution set of the inequality −3x + 4 ≥ 10 is the interval (−∞,−2].
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Determine whether the statement is true or false. If false, describe how the statement might be changed to make it true. 7. The only solution to |3x − 4| = 2 is x = 2/3.
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Determine whether the statement is true or false. If false, describe how the statement might be changed to make it true. 8. The solution set of the inequality |x − 4| ≤ 3 is the interval [1, 7].
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Determine whether the statement is true or false. If false, describe how the statement might be changed to make it true. 9. If |x − 5| = 3, then the distance from x to 5 is 3.
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