Find the equation of the line through the y-intercept?

Question

Find the equation of the line through the y-intercept of $y = x^{4} - 8 x^{5} - 7 + 6 x^{2}$ $y=x^4-8x^5-7+6x^2$ and the x intercept of y=3x-9?

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Answer 1 · 2017-11-19T02:25:18

$y = \frac{7}{3} x - 7$ $y=7/3 x - 7$

The $y$ $y$ -intercept of $y = x^{4} - 8 x^{5} - 7 + 6 x^{2}$ $y=x^4-8x^5-7+6x^2$ occurs where $x = 0$ $x=0$ .

$y = {(0)}^{4} - 8 {(0)}^{5} - 7 + 6 {(0)}^{2} = - 7$ $y=(0)^4-8(0)^5-7+6(0)^2=-7$

The $x$ $x$ -intercept of $y = 3 x - 9$ $y=3x-9$ occurs where $y = 0$ $y=0$ .

$0 = 3 x - 9$ $0=3x-9$

$9 = 3 x$ $9=3x$

$\frac{9}{3} = \frac{3 x}{3}$ $9/3=(3x)/3$

$x = 3$ $x=3$

So the question is what line goes through the points $(0, - 7)$ $(0,-7)$ and $(3, 0)$ $(3, 0)$ . The equation for the slope of a line is given as

$m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}$ $m = (y_2-y_1)/(x_2-x_1)$

$m = \frac{0 - (- 7)}{3 - 0} = \frac{7}{3}$ $m=(0-(-7))/(3-0)=7/3$

The $y$ $y$ -intercept $b$ $b$ can be determined by plugging in the slope and one of the two points (it doesn't matter which one).

$y = m x + b$ $y=mx+b$

$- 7 = \frac{7}{3} (0) + b$ $-7=7/3(0) + b$

$b = - 7$ $b=-7$

With $m = \frac{7}{3}$ $m=7/3$ and $b = - 7$ $b=-7$ , the equation of a line becomes

$y = m x + b$ $y=mx+b$

$y = \frac{7}{3} x - 7$ $y=7/3 x - 7$