Question

What are the x and y intercepts of the equation?y= log (3x + 4) + 2

Answer 1

`y=log(3x+4)+2`

To find the x-intercept we make the value of y=0, then,

`0=log(3x+4)+2`

clear the equation for x

`\begin{gathered} -2=log(3x+4) \\ write\text{ }the\text{ }equation\text{ }in\text{ }exponential\text{ }form \\ 10^(-2)=3x+4 \\ (1)/(10^2)=3x+4 \\ solve\text{ }for\text{ }x \\ (1)/(100)=3x+4 \\ \\ (1)/(100)-4=3x \\ \\ -(399)/(100)=3x \\ \\ x=-(399)/(100)*(1)/(3) \\ \\ x=-(133)/(100)=-1.33 \end{gathered}`

The x-intercept is -1.33

To find the y-intercept we make the value of x=0, then,

`\begin{gathered} y=log(3(0)+4)+2 \\ y=log(4)+3 \\ using\text{ }exponential\text{ }form \\ y=log(2^2)+2 \\ y=2log(2)+2 \\ y\cong2.602 \end{gathered}`

Answer:

The intercepts are:

y-intercept is y=2.602

x-intercept is x=-1.33