Simple Functional Relationships - ISEE Upper Level: Quantitative Reasoning

$5$. Compute using the formula for average rate of change over the interval from 1 to 4.

$\frac{f(b)-f(a)}{b-a}$. The average rate of change is the slope of the secant line between $x=a$ and $x=b$.

$x$ is the input; $y$ is the output. In the notation $y = f(x)$, $x$ serves as the independent variable or input, while $y$ is the dependent variable or output.

$13$. Substitute $x=5$ into the linear function to compute the output as $3(5) - 2$.

$5$. Evaluate the quadratic function by substituting $x = -2$ to get $(-2)^2 + 1$.

$4$. Solve the equation $2x + 7 = 15$ by subtracting 7 and dividing by 2 to find the input.

$m=\frac{\Delta y}{\Delta x}$. The slope $m$ represents the ratio of the change in $y$ to the change in $x$ for a linear function.

$2$. Calculate the slope using the formula $\frac{y_2 - y_1}{x_2 - x_1}$ with the given points.

$9$. In slope-intercept form, the constant term is the $y$-coordinate where the line crosses the $y$-axis.

The value of $b$. The $y$-intercept occurs when $x=0$, so it equals $b$ in the slope-intercept form.

$y=kx$. A proportional relationship passes through the origin with slope equal to the constant $k$.

$\frac{5}{2}$. Determine $k$ by dividing $y$ by $x$ from the given point in the proportional relationship.

$8$. Evaluate the exponential function by raising the base 2 to the power of 3.

$3$. The absolute value function computes the distance from $x$ to 5, yielding a non-negative result.

$f(x)=3x+1$. Find the slope from the points, then use point-slope form to derive the linear equation.

$(3,-2)$. The ordered pair lists the input first and the corresponding output second.

$7$. First evaluate $g(2)$, then substitute that result into $f$ for composition.

A relation where each input has exactly one output. A function ensures that for every input value in the domain, there is precisely one corresponding output value.

$f^{-1}(x)=x-7$. The inverse undoes the operation by subtracting 7 to recover the original input.

$y\ge -4$. The parabola shifts down by 4 units, making the minimum value -4 at the vertex.

$y\ge 0$. The square of any real number is non-negative, so outputs are bounded below by zero.

$x\ge -5$. For real square roots, the expression inside must be non-negative, setting the domain boundary.

$x\ne 3$. The domain excludes values that make the denominator zero to avoid undefined results.

$x^2+4x+4$. Substitute $x+2$ into the function and expand the binomial to simplify.