Chapter 1.2, Problem 29E

To determine

ToDetermineThe $x-\text{intercept}$ and $y-\text{intercept}$ of the graph of the equation $y=2x^3-4x^2$

Answer to Problem 29E

The $x-\text{intercepts}$ of the graph are the points $\left(x,y\right)=\left(0,0\right)$ and $\left(x,y\right)=\left(2,0\right)$

The $y-\text{intercept}$ of the graph is the point $\left(x,y\right)=\left(0,0\right)$

Explanation of Solution

Given:

The equation $y=2x^3-4x^2$

Concept Used:

The $x-\text{intercept}$ is a point on the graph where the graph intersects $x-\text{axis}$ i.e., where the $y$ coordinate is zero.

The $y-\text{intercept}$ is a point on the graph where the graph intersects $y-\text{axis}$ i.e., where the $x$ coordinate is zero.

Calculation:

Given the equation $y=2x^3-4x^2$

For $x-\text{intercept}$ , substituting $y=0$

i.e., $y=2x^3-4x^2$

i.e., $\begin{array}{c}0=2x^3-4x^2\\ 0=2x^2\left(x-2\right)\end{array}$

i.e., $2x^2=0$

and

  $x-2=0$

i.e., $x=0$

and

  $x=0$

and

  $x=2$

Therefore, $x-\text{intercepts}$ of the graph are the points $\left(x,y\right)=\left(0,0\right)$ and $\left(x,y\right)=\left(2,0\right)$

For $y-\text{intercept}$ , substituting $x=0$

i.e., $y=2x^3-4x^2$

i.e., $\begin{array}{l}y=2{\left(0\right)}^3-4{\left(0\right)}^2\\ y=0-0\\ y=0\end{array}$

Therefore, $y-\text{intercept}$ of the graph is the point $\left(x,y\right)=\left(0,0\right)$

Conclusion:

The $x-\text{intercepts}$ of the graph are the points $\left(x,y\right)=\left(0,0\right)$ and $\left(x,y\right)=\left(2,0\right)$

The $y-\text{intercept}$ of the graph is the point $\left(x,y\right)=\left(0,0\right)$