Open Practice Tests

Nonlinear function features and transformations — Topic practice

5 exercises · 2 automatically checked · 3 written self-checks · Untimed

Instructions and review guidance

2 questions · Shuffled order · Untimed practice

Work through the set before checking solutions. Guide question labels match the source lesson references. 3 written responses appear below the question bank, with model answers for self-review; these are not automatically graded.

Guide question labels stay the same when the order changes. Check answers when you are ready to review.

Written self-checks are below the question bank and are excluded from the automatic score.

Written responses are self-review exercises.

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The drill

Question 1

  • Guide question E6.1

If f(x) = −x2 + 3x + 4, find f(−2).

Enter an integer, decimal or fraction without units. Use up to 5 characters, or 6 including a leading minus.

Question 5

  • Guide question E6.5

If f(x) = a(x − 2)2 + 3 and f(4) = 15, find a.

Enter an integer, decimal or fraction without units. Use up to 5 characters, or 6 including a leading minus.

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Written self-checks

Write your complete response before opening the model answer. Compare both your answer and your reasoning.

Guide question E6.2

The graph of f contains (3, 7). Find the corresponding point on g(x) = f(x + 5) − 2.

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(−2, 5)

The known point means f(3) = 7. For g(x) = f(x + 5) − 2, set the inside input equal to 3: x + 5 = 3, so x = −2. The new output is 7 − 2 = 5. Hence (−2, 5) must lie on the new graph. A plus sign inside shifts points left. Review 6.05.

Guide question E6.3

Find the real domain of h(x) = 2x − 6.

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[3, ∞)

Real outputs require a nonnegative radicand:

2x − 6 ≥ 0 x ≥ 3.

The endpoint is included because 0 is defined. The range would describe outputs, not inputs; this question asks for the domain. Review 6.09.

Guide question E6.4

Find the range of f(x) = 5 − (x − 1)2 when 0 ≤ x ≤ 4.

Show model answerHide model answer for question E6.4

[−4, 5] For f(x) = 5 − (x − 1)2 on [0, 4], the vertex input 1 lies in the domain and gives the maximum 5. Evaluate both endpoints: f(0) = 4 and f(4) = −4. The minimum is −4, and the continuous graph takes every value between the extremes. Thus the range is [−4, 5]. Review 6.10.