What are the intercepts of $- 7 y = 3 y - 2 (x - 9) - x^{2}$ $-7y=3y-2(x-9) -x^2$ ?

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Answer 1 · 2018-06-06T00:28:28

$x = - 10 \pm \sqrt{19}$ $x=-10+-sqrt19$

$y = - \frac{9}{5}$ $y=-9/5$

To find the y-intercepts set x=0 and solve for y:

$- 7 y = 3 y - 2 (x - 9) - x^{2}$ $-7y=3y-2(x-9) -x^2$

$- 7 y = 3 y - 2 (0 - 9) - 0^{2}$ $-7y=3y-2(0-9) -0^2$

$- 7 y = 3 y - 2 (- 9)$ $-7y=3y-2(-9)$

$- 7 y = 3 y + 18$ $-7y=3y+18$

$- 10 y = 18$ $-10y=18$

$y = - \frac{9}{5}$ $y=-9/5$

To find the x-intercept(s) if they exist set y=0 and solve for x:

$- 7 y = 3 y - 2 (x - 9) - x^{2}$ $-7y=3y-2(x-9) -x^2$

$- 7 (0) = 3 (0) - 2 (x - 9) - x^{2}$ $-7(0)=3(0)-2(x-9) -x^2$

$0 = - 2 (x - 9) - x^{2}$ $0=-2(x-9) -x^2$

$0 = - x^{2} - 2 (x - 9)$ $0=-x^2-2(x-9)$

$0 = - x^{2} - 2 x + 18$ $0=-x^2-2x+18$

$0 = x^{2} + 2 x - 18$ $0=x^2+2x-18$

You will need to complete the square or use the quadratic equation to find these roots:

$x = - 10 \pm \sqrt{19}$ $x=-10+-sqrt19$

graph{-7y=3y-2(x-9) -x^2 [-20.58, 19.42, -4.8, 15.2]}

What are the intercepts of −7y=3y−2(x−9)−x2-7y=3y-2(x-9) -x^2?