Graphic Function Practice Questions

The expression under a square root must be greater than or equal to zero. Set the radicand \(\geq 0\) and solve:

\(x – 3 \geq 0\)

\(x \geq 3\)

Choice B uses a strict inequality (\(\gt\) instead of \(\geq\)), which would exclude \(x = 3\). However, \(f(3) = \sqrt{0} = 0\) is defined, so \(x = 3\) must be included in the domain.

Transformations of a parent function \(f(x)\) follow these rules:

• Horizontal shift: \(f(x – h)\) shifts the graph \(h\) units to the right. \(f(x + h)\) shifts it to the left.
• Vertical shift: \(f(x) + k\) shifts the graph \(k\) units up. \(f(x) – k\) shifts it down.

Shifting 3 units right replaces \(x\) with \((x – 3)\), and shifting 2 units up adds 2:

\(g(x) = (x – 3)^2 + 2\)

Choice A shifts left instead of right. Remember: the horizontal shift is counterintuitive — a minus sign moves right, and a plus sign moves left.

To test whether a function is even or odd, evaluate \(f(-x)\) and compare it to \(f(x)\):

\(f(-x) = (-x)^3 – (-x)\)\(\:= -x^3 + x\)\(\:= -(x^3 – x)\)\(\:= -f(x)\)

Since \(f(-x) = -f(x)\), the function is odd. The rules are:

• Even: \(f(-x) = f(x)\) — symmetric about the \(y\)-axis
• Odd: \(f(-x) = -f(x)\) — symmetric about the origin

If neither condition is met, the function is neither even nor odd.

Start with the parent function \(|x|\), which always produces values \(\geq 0\). Then consider the transformations:

• The negative sign in \(-|x|\) reflects the graph over the \(x\)-axis, so \(-|x| \leq 0\).
• Adding 5 shifts everything up by 5, so \(-|x| + 5 \leq 5\).

The maximum value of the function is 5 (occurring at \(x = 0\)), and the function decreases without bound as \(|x|\) increases. Therefore the range is \(y \leq 5\).

Choice A describes the range of \(|x| + 5\) (without the reflection), which opens upward instead of downward.

To identify the parent function, look at the core operation being performed on \(x\) — ignoring any shifts, stretches, or reflections. The function \(g(x) = 3\sqrt{x + 1} – 4\) is a transformed version of the square root function \(f(x) = \sqrt{x}\).

The transformations applied are:

• Horizontal shift 1 unit left (\(x + 1\) inside the radical)
• Vertical stretch by a factor of 3 (coefficient of 3)
• Vertical shift 4 units down (\(– 4\) outside)

Break the expression \(-f(x – 2) + 3\) into its individual transformations, reading from the inside out:

• \((x – 2)\): Horizontal shift 2 units to the right.
• Negative sign on \(f\): Reflection over the \(x\)-axis (flips the graph upside down).
• \(+ 3\): Vertical shift 3 units up.

Choice A incorrectly interprets \((x – 2)\) as a left shift. Choice B confuses reflecting over the \(x\)-axis (negating the output) with reflecting over the \(y\)-axis (negating the input).

The \(x\)-intercepts are the points where \(f(x) = 0\). Set the function equal to zero and factor:

\(x^2 – 4x – 5 = 0\)

Find two numbers that multiply to −5 and add to −4:

\((x – 5)(x + 1) = 0\)

So \(x = 5\) and \(x = -1\). The \(x\)-intercepts are the points \((-1, 0)\) and \((5, 0)\).

Choice C gives points on the \(y\)-axis rather than the \(x\)-axis. Remember: \(x\)-intercepts always have a \(y\)-coordinate of 0, and \(y\)-intercepts always have an \(x\)-coordinate of 0.

Apply each transformation one at a time to the parent function \(f(x) = |x|\):

Step 1: Reflect over the \(x\)-axis. Negate the entire function:

\(-|x|\)

Step 2: Shift 4 units left. Replace \(x\) with \((x + 4)\):

\(-|x + 4|\)

Choice A shifts right instead of left. Choice D shifts down 4 units instead of left 4 units. Subtracting 4 outside the absolute value is a vertical shift, not a horizontal one.

The average rate of change of a function from \(x = a\) to \(x = b\) is the slope of the line connecting the two points:

\(\dfrac{f(b) – f(a)}{b – a}\)

First, evaluate the function at each endpoint:

\(f(1) = (1)^2 + 2(1) = 3\)

\(f(4) = (4)^2 + 2(4) = 24\)

Now compute the average rate of change:

\(\dfrac{f(4) – f(1)}{4 – 1} = \dfrac{24 – 3}{3} = \dfrac{21}{3} = 7\)

Choice D gives the change in \(f(x)\) alone (21) without dividing by the change in \(x\).

For a piecewise function, first determine which piece applies based on the \(x\)-value. Since \(x = 3\) satisfies \(0 \leq x \leq 3\), use the second piece:

\(f(3) = 2(3) – 5 = 6 – 5 = 1\)

Choice C uses the third piece (\(4x = 12\)), but that piece only applies when \(x \gt 3\), not when \(x = 3\). Pay close attention to whether the boundary uses \(\lt\) / \(\gt\) (exclusive) or \(\leq\) / \(\geq\) (inclusive).

Compare \(g(x) = 2|x + 3|\) to the parent function \(f(x) = |x|\) by identifying each change:

• The coefficient 2: Since it multiplies the entire output, it is a vertical stretch by a factor of 2. This makes the V-shape narrower.
• \((x + 3)\) inside the absolute value: This shifts the graph 3 units to the left.

Choice A gets the shift direction wrong. Choice C confuses a vertical stretch (multiplying outside) with a horizontal stretch (multiplying the input inside).

The \(y\)-intercept is the point where \(x = 0\). To find it, substitute \(x = 0\) into each function:

• Choice A: \(f(0) = 3(0)^2 + 2(0) – 7 = -7\)
• Choice B: \(f(0) = (0)^2 – 7(0) = 0\)
• Choice C: \(f(0) = -7(0) + 1 = 1\)
• Choice D: \(f(0) = (0)^3 + 7 = 7\)

Only choice A produces a \(y\)-intercept of −7. Note that in a polynomial written in standard form, the \(y\)-intercept is always the constant term.

An inverse function \(f^{-1}\) reverses the input and output of the original function. If \(f\) maps 3 to 8, then \(f^{-1}\) maps 8 back to 3:

\(f(3) = 8 \implies f^{-1}(8) = 3\)

In other words, the inverse function swaps the \(x\)– and \(y\)-values. Graphically, the graph of \(f^{-1}(x)\) is a reflection of the graph of \(f(x)\) over the line \(y = x\).

Choice C confuses \(f^{-1}(x)\) (the inverse function) with \(\frac{1}{f(x)}\) (the reciprocal). These are not the same thing.

The domain excludes any values that make the denominator zero. Set the denominator equal to zero:

\(x^2 – 25 = 0\)

This is a difference of squares:

\((x + 5)(x – 5) = 0\)

So \(x = 5\) and \(x = -5\) must both be excluded from the domain.

Choice A confuses \(x^2 = 25\) with \(x = 25\). Choice B only excludes one of the two values — don’t forget that \(x^2 = 25\) has two solutions.

When the input \(x\) is multiplied by a constant inside the function, it creates a horizontal transformation. For \(g(x) = f(cx)\):

• If \(|c| \gt 1\): The graph is compressed horizontally by a factor of \(\frac{1}{c}\).
• If \(0 \lt |c| \lt 1\): The graph is stretched horizontally by a factor of \(\frac{1}{c}\).

Here \(c = 2\), so the graph is compressed horizontally by a factor of \(\frac{1}{2}\). Every \(x\)-value is halved, making the graph appear “squeezed” toward the \(y\)-axis.

This is another counterintuitive transformation: multiplying \(x\) by a number greater than 1 makes the graph narrower, not wider. Choice C confuses a horizontal compression (inside the function) with a vertical stretch (outside the function).