NBT Mathematics: Functions and Graphs Flashcards
7 cards from real NBT practice questions. Tap to flip, then mark Knew It or Still Learning — missed cards come back until you master them.
Read the first 7 NBT Mathematics: Functions and Graphs flashcards as text
The graph of y = ax + b passes through (0, 4) and (2, 0). What are the values of a and b?
Answer: a = −2, b = 4
y-intercept gives b = 4; gradient = (0 − 4) ÷ (2 − 0) = −2, so a = −2.
Which of the following is the equation of a parabola with vertex at (2, −3) and passing through (0, 1)?
Answer: y = (x − 2)² − 3
Vertex (2, −3) gives y = a(x − 2)² − 3. Substituting (0, 1): 1 = 4a − 3 → a = 1. So y = (x − 2)² − 3.
The function p(x) = 2^x. The graph of q(x) is p(x) shifted 3 units upward. What is q(x)?
Answer: q(x) = 2^x + 3
A vertical shift of 3 units upward adds 3 to the function: q(x) = 2^x + 3.
The domain of f(x) = √(x − 5) is:
Answer: x ≥ 5
The expression under the square root must be ≥ 0: x − 5 ≥ 0 → x ≥ 5.
A graph shows f(x) crossing the x-axis at x = −3 and x = 2, with a y-intercept at y = −6. Which function fits?
Answer: f(x) = (x + 3)(x − 2)
Roots at −3 and 2 → f(x) = a(x + 3)(x − 2). At x = 0: a(3)(−2) = −6 → −6a = −6 → a = 1.
What is the y-intercept of the exponential function f(x) = 5 · (0.4)^x?
Answer: (0, 5)
f(0) = 5 · (0.4)^0 = 5 · 1 = 5, so y-intercept is (0, 5).
For f(x) = x² − 6x + 5, find the vertex by completing the square.
Answer: (3, −4)
f(x) = (x − 3)² − 9 + 5 = (x − 3)² − 4; vertex = (3, −4).