Linear, Quadratic & Polynomial Functions (College Board AP® Precalculus): Multiple Choice Questions

$f$

$g$

$h$

$k$

$x^3-2x^2-5x-6$

$x^3-2x^2-5x+6$

$x^3-2x^2+5x+6$

$x^3+2x^2-5x+6$

$-80$

$4$

$20$

$40$

$\left[0,5\right]$

$\left[-3,5\right]$

$\left(-\infty ,-3\right]\cup \left[0,5\right]$

$\left[-3,0\right]\cup \left[5,\infty \right)$

$\underset{x\to -\infty }{\lim }g\left(x\right)=-\infty$ and $\underset{x\to \infty }{\lim }g\left(x\right)=-\infty$

$\underset{x\to -\infty }{\lim }g\left(x\right)=\infty$ and $\underset{x\to \infty }{\lim }g\left(x\right)=\infty$

$\underset{x\to -\infty }{\lim }g\left(x\right)=-\infty$ and $\underset{x\to \infty }{\lim }g\left(x\right)=\infty$

$\underset{x\to -\infty }{\lim }g\left(x\right)=\infty$ and $\underset{x\to \infty }{\lim }g\left(x\right)=-\infty$

$p$ has exactly three distinct real zeros.

$p$ has exactly two distinct real zeros.

$p$ has exactly one distinct real zero and no non-real zeros.

$p$ has exactly one distinct real zero and two non-real zeros.

$h$ is best modeled by a linear function, because the rate of change over consecutive equal-length input-value intervals is constant.

$h$ is best modeled by a linear function, because the change in the average rates of change over consecutive equal-length input-value intervals is constant.

$h$ is best modeled by a quadratic function, because the rate of change over consecutive equal-length input-value intervals is constant.

$h$ is best modeled by a quadratic function, because the change in the average rates of change over consecutive equal-length input-value intervals is constant.

The graph of $k$ has three $x$-intercepts.

$-1.23+0.56i$ is a zero of $k$.

The equation $k\left(x\right)=0$ has four real solutions.

The graph of $k$ is tangent to the $x$-axis at $x=12.458$.

The sign of the leading term of $p$ is positive, and the degree of the leading term is even; therefore $\underset{x\to -\infty }{\lim }p\left(x\right)=\infty$ and $\underset{x\to \infty }{\lim }p\left(x\right)=\infty$.

The sign of the leading term of $p$ is positive, and the degree of the leading term is odd; therefore $\underset{x\to -\infty }{\lim }p\left(x\right)=-\infty$ and $\underset{x\to \infty }{\lim }p\left(x\right)=\infty$.

The sign of the leading term of $p$ is negative, and the degree of the leading term is even; therefore $\underset{x\to -\infty }{\lim }p\left(x\right)=-\infty$ and $\underset{x\to \infty }{\lim }p\left(x\right)=-\infty$.

The sign of the leading term of $p$ is negative, and the degree of the leading term is odd; therefore $\underset{x\to -\infty }{\lim }p\left(x\right)=\infty$ and $\underset{x\to \infty }{\lim }p\left(x\right)=-\infty$.

$p(-5)=3$ and $p(-5)$ is a relative maximum

$p(-5)=-3$ and $p(-5)$ is a relative maximum

$p(-5)=3$ and $p(-5)$ is a relative minimum

$p(-5)=-3$ and $p(-5)$ is a relative minimum

$0.5(x+4)(x-2)(x-6)$

$0.5(x+4{)}^2(x-2)(x-6)$

$0.5(x+4)(x-2{)}^2(x-6)$

$0.5(x+4)(x-2{)}^2(x+6)$

$t=1.800$

$t=4.609$

$t=11.391$

$t=16$

$h$ is best modeled by a linear function, because the rate of change over consecutive equal-length input-value intervals is constant.

$h$ is best modeled by a linear function, because the change in the average rates of change over consecutive equal-length input-value intervals is constant.

$h$ is best modeled by a quadratic function, because the rate of change over consecutive equal-length input-value intervals is constant.

$h$ is best modeled by a quadratic function, because the change in the average rates of change over consecutive equal-length input-value intervals is constant.

$t=16$

$t=11.162$

$t=4.838$

$t=2.500$

Linear

Quadratic

Cubic

Quartic

$p$ is even, because $p\left(-x\right)=p\left(x\right)$ for all real values of $x$.

$p$ is neither even nor odd, because the graph of $p$ does not pass through the origin.

$p$ is neither even nor odd, because $p\left(-x\right)=\left(-x-2\right)\left(-x+2\right)\left(-x^2+1\right)$, which is not equal to $p\left(x\right)$.

$p$ is odd, because $p\left(-x\right)=-p\left(x\right)$ for all real values of $x$.

$\left(-\infty ,-2\right)\cup \left(1,\infty \right)$

$\left(-2,1\right)$

$\left(-2,1\right)\cup \left(1,\infty \right)$

$\left(-2,\infty \right)$

The sign of the leading term of $f$ is negative, and the degree of the leading term of $f$ is even; therefore, $\underset{x\to -\infty }{\lim }f\left(x\right)=-\infty$ and $\underset{x\to \infty }{\lim }f\left(x\right)=-\infty$.

The sign of the leading term of $f$ is negative, and the degree of the leading term of $f$ is odd; therefore, $\underset{x\to -\infty }{\lim }f\left(x\right)=\infty$ and $\underset{x\to \infty }{\lim }f\left(x\right)=-\infty$.

The sign of the leading term of $f$ is positive, and the degree of the leading term of $f$ is even; therefore, $\underset{x\to -\infty }{\lim }f\left(x\right)=\infty$ and $\underset{x\to \infty }{\lim }f\left(x\right)=\infty$.

The sign of the leading term of $f$ is positive, and the degree of the leading term of $f$ is odd; therefore, $\underset{x\to -\infty }{\lim }f\left(x\right)=-\infty$ and $\underset{x\to \infty }{\lim }f\left(x\right)=\infty$.

$x=a$

$x=\frac{a+b}{2}$

$x=b$

The location cannot be determined.

$-1080$

$-720$

$-540$

$1080$

$1$

$4$

$5$

$7$

$-3$

$-1$

$5$

$7$