12 multiple-choice questions, progressively harder.
The graph of fff passes through (−2,7)(-2, 7)(−2,7), (−1,0)(-1, 0)(−1,0), (0,−5)(0, -5)(0,−5), (1,−8)(1, -8)(1,−8), (2,−9)(2, -9)(2,−9), (3,−8)(3, -8)(3,−8), and (4,−5)(4, -5)(4,−5). What is the lowest output shown, and at which input?
Solution
Correct answer: A
The lowest output is the smallest second coordinate among the points.
minimum listed output=−9 at (2,−9)\text{minimum listed output} = -9 \text{ at } (2, -9)minimum listed output=−9 at (2,−9)
So the minimum shown is −9-9−9, reached at x=2x = 2x=2.
For f(x)=x2−4x+3f(x) = x^2 - 4x + 3f(x)=x2−4x+3, find both x-intercepts and the y-intercept.
Correct answer: B
Solve f(x)=0f(x) = 0f(x)=0 by factoring, and evaluate f(0)f(0)f(0).
x2−4x+3=(x−1)(x−3)=0 ⇒ x=1,3;f(0)=3x^2 - 4x + 3 = (x - 1)(x - 3) = 0 \;\Rightarrow\; x = 1, 3; \qquad f(0) = 3x2−4x+3=(x−1)(x−3)=0⇒x=1,3;f(0)=3
So the x-intercepts are 111 and 333, and the y-intercept is (0,3)(0, 3)(0,3).
The graph of fff is shown. What is the minimum value of fff, and where does it occur?
Correct answer: D
The minimum value is the lowest height the curve reaches, at the bottom of the parabola.
lowest point (1,−4) ⇒ minimum −4\text{lowest point } (1, -4) \;\Rightarrow\; \text{minimum } -4lowest point (1,−4)⇒minimum −4
The minimum output is −4-4−4, reached at the input x=1x = 1x=1.
The graph of fff has x-intercepts at −3-3−3 and 111 and opens upward (a parabola). On which interval is fff negative (below the x-axis)?
An upward parabola dips below the axis exactly between its two x-intercepts.
−3<x<1 ⇒ f(x)<0-3 < x < 1 \;\Rightarrow\; f(x) < 0−3<x<1⇒f(x)<0
Outside that interval the curve is above the axis.
The graph of fff is shown. On which interval is fff decreasing?
Correct answer: C
The downward parabola rises to its peak at x=1x = 1x=1, then falls.
falls for x>1 ⇒ decreasing for x>1\text{falls for } x > 1 \;\Rightarrow\; \text{decreasing for } x > 1falls for x>1⇒decreasing for x>1
To the left of the peak it is increasing.
For f(x)=x2−6x+8f(x) = x^2 - 6x + 8f(x)=x2−6x+8, at which input does the graph reach its lowest point (its vertex)?
The lowest point of an upward parabola sits at x=−b2ax = -\dfrac{b}{2a}x=−2ab, midway between the roots.
x=−−62(1)=3x = -\frac{-6}{2(1)} = 3x=−2(1)−6=3
The vertex, and the minimum, occur at x=3x = 3x=3.
The point (k,7)(k, 7)(k,7) lies on the graph of f(x)=2x+1f(x) = 2x + 1f(x)=2x+1. What is kkk?
Since the point is on the graph, its coordinates satisfy y=f(x)y = f(x)y=f(x).
7=2k+1 ⇒ k=37 = 2k + 1 \;\Rightarrow\; k = 37=2k+1⇒k=3
So the point is (3,7)(3, 7)(3,7).
The graph of fff passes through (−3,−4)(-3, -4)(−3,−4) and (1,8)(1, 8)(1,8). What is the slope of the straight line through these two points?
Divide the change in output by the change in input between the two points.
8−(−4)1−(−3)=124=3\frac{8 - (-4)}{1 - (-3)} = \frac{12}{4} = 31−(−3)8−(−4)=412=3
That is the slope of the segment joining them.
The graph of y=f(x)y = f(x)y=f(x) and the horizontal line y=−1y = -1y=−1 do not meet at all. What can you conclude?
Meeting points of the line y=−1y = -1y=−1 and the graph are the solutions of f(x)=−1f(x) = -1f(x)=−1, and there are none.
no meeting ⇒ f(x)=−1 has no solution\text{no meeting} \;\Rightarrow\; f(x) = -1 \text{ has no solution}no meeting⇒f(x)=−1 has no solution
So the output −1-1−1 is never produced, meaning −1-1−1 is outside the range.
For f(x)=x2−1f(x) = x^2 - 1f(x)=x2−1, which describes where the graph is above the x-axis (where f(x)>0f(x) > 0f(x)>0)?
The graph is above the axis outside its two x-intercepts at −1-1−1 and 111.
x2−1>0 ⇒ x<−1 or x>1x^2 - 1 > 0 \;\Rightarrow\; x < -1 \text{ or } x > 1x2−1>0⇒x<−1 or x>1
Between −1-1−1 and 111 the curve dips below the axis.
The graph of fff is increasing for x<0x < 0x<0 and decreasing for x>0x > 0x>0. Which point is a maximum?
The graph rises up to x=0x = 0x=0 and falls after it, so x=0x = 0x=0 is the top of a hill.
rise then fall at x=0 ⇒ maximum at x=0\text{rise then fall at } x = 0 \;\Rightarrow\; \text{maximum at } x = 0rise then fall at x=0⇒maximum at x=0
That turning point is the largest output in its neighborhood.
A graph passes the vertical line test and also meets some horizontal line at two points. Which is true?
Passing the vertical line test makes it a function; meeting a horizontal line twice means one output is shared by two inputs.
one output per input, but one output from two inputs\text{one output per input, but one output from two inputs}one output per input, but one output from two inputs
That is allowed: a function may repeat outputs, it just may not give one input two outputs.
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