Question in-qua
2.1 Nonlinear Functions - Quadratic graphs and vertex form
0:00
- In quadrant 3, the curve rises to its maximum at point (0 comma negative 3).
- In quadrant 4, the curve then falls.
The graph of the polynomial function , where , is shown. The y-intercept of the graph is . What is the value of ?
Enter your answer:
In quadrant 3, the curve rises to its maximum at point (0 comma negative 3). In
Medium-difficulty · SAT Math · Nonlinear Functions — Quadratic graphs and vertex form. Read the question above, select your answer, and check the full explanation below to understand exactly why the correct choice works.
Answer explanation
The correct answer is . The y-intercept of the graph of a function in the xy-plane is the point where the graph crosses the y-axis. The graph of the polynomial function shown crosses the y-axis at the point . It's given that the y-intercept of the graph is . Thus, the value of is .
More Nonlinear Functions practice questions
- f x = 4 x 2 - 50 x + 126 The given equation
- The function p is defined by p n = 7 n 3 . What
- The parabola opens downward. The vertex is at point (4 comma 9). The parabola
- f x = x + 7 2 + 4 The function f is defined
- The function f w = 6 w 2 gives the area of a rectangle,
- The exponential function g is defined by g x = 19 · a x
- q ( x ) = 32 ( 2 x ) Which table gives three
- A quadratic function models the height, in feet, of an object above the ground
Browse all Passport to Advanced Math practice questions or return to the full SAT question bank.