# How do you find the x and y intercepts for -4x-7y+7=0?

Mar 2, 2018

I tried this:

#### Explanation:

You need to set $x = 0$ and $y = 0$ in your expression.

Let us find the $x$ intercept; we need to, basically, stay on the $x$ axis so that $y$ has to be zero (otherwise we are not on the $x$ axis).
So, we get:

$y = 0$
$- 4 x - 7 \cdot \textcolor{red}{0} + 7 = 0$
$- 4 x = - 7$
$x = \frac{7}{4}$
So the $x -$axis intercept will be:
$\left(\frac{7}{4} , 0\right)$

Now for the $y$ intercept we need $x$ to be zero and we get:
$x = 0$
$- 4 \cdot \textcolor{red}{0} - 7 y + 7 = 0$
$- 7 y = - 7$
$y = \frac{7}{7} = 1$
So the $y -$axis intercept will be:
$\left(0 , 1\right)$

Let us check and "see" our intercepts graphically:
graph{-(4/7)x+1 [-10, 10, -5, 5]}

Mar 2, 2018

$\text{x-intercept "=7/4," y-intercept } = 1$

#### Explanation:

$\text{to find the intercepts that is where the graph crosses}$
$\text{the x an d y axes}$

• " let x = 0, in the equation for y-intercept"

• " let y = 0, in the equation for x-intercept"

$- 4 x - 7 y + 7 = 0$

$\Rightarrow - 4 x - 7 y = - 7$

$x = 0 \Rightarrow 0 - 7 y = - 7 \Rightarrow y = 1 \leftarrow \textcolor{red}{\text{y-intercept}}$

$y = 0 \Rightarrow - 4 x + 0 = - 7 \Rightarrow x = \frac{7}{4} \leftarrow \textcolor{red}{\text{x-intercept}}$
graph{(y+4/7x-1)((x-7/4)^2+(y-0)^2-0.04)((x-0)^2+(y-1)^2-0.04)=0 [-10, 10, -5, 5]}