Question

Graph the function below. Plot 5 points on the graph of the function, one point with x equals zero, two points with negative x values, and two points with positive x values.

Answer 1

Solution:

Let the function:

$y\text{ = }\frac{-11}{8}x^3$

Since it is a cubic function, its graph is

now, the points on the graph, that are requested are obtained as follows:

For x= 0 then

$y=f(0)=\frac{-11}{8}(0^3)=0$

thus, the point with x equals zero is:

$A\text{ = (0,0)}$

now, for x = -1 then:

$y=f(-1)=\frac{-11}{8}(-1^3)=\frac{11}{8}$

thus, a point with a negative x value is:

$B\text{ =(-1,}\frac{11}{8}\text{)}$

also, for x = -2, we get:

$y=f(-2)=\frac{-11}{8}(-2^3)=11$

then, another point with a negative x value is:

$C\text{ = (}-2,11\text{)}$

now, if x = 1, then

$y=f(1)=\frac{-11}{8}(1^3)=\frac{-11}{8}$

thus, a point with a positive x value is:

$D\text{ = (1,-}\frac{11}{8}\text{)}$

also, if x = 2, we get:

$y=f(2)=\frac{-11}{8}(2^3)=-11$

then, another point with a positive x value is:

$E\text{ = }(2,-11)$

On the graph, these points are:

So that, we can conclude that the correct answer is:

The graph of the function is:

one point with x equals zero is:

$A\text{ = (0,0)}$

two points with negative x values are:

$B\text{ =(-1,}\frac{11}{8}\text{)}$

$C\text{ = (}-2,11\text{)}$

two points with positive x values are:

$D\text{ = (1,-}\frac{11}{8}\text{)}$

$E\text{ = }(2,-11)$