How do you graph $(x - 1) (x + 4) - y = 0$ $(x-1)(x+4)-y=0$ ?

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Answer 1 · 2018-02-01T19:13:19

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$(x - 1) (x + 4) - y = 0$ $(x-1)(x+4)-y=0$ can be simpified as

$x^{2} + 3 x - 4 = y$ $x^2+3x-4=y$

now since we have a quadratic, label a, b, and c.

a=1
b=3
c=4 -- this is the y-intercept

now use the formula $\frac{- b}{2 a}$ $(-b)/(2a)$ to find the symmetrical line (which is imaginary)

$\frac{- 3}{2 (1)}$ $\frac{-3}{2(1)}$ = $\frac{- 3}{2}$ $\frac{-3}{2}$

now use that to substitute x, which will be ur $y$ $y$

Btw, x is $- \frac{3}{2}$ $-3/2$

now, ${(\frac{- 3}{2})}^{2} + 3 (\frac{- 3}{2}) - 4$ $(\frac{-3}{2})^2+3(\frac{-3}{2})-4$

= $\frac{- 25}{4}$ $\frac{-25}{4}$

$(\frac{- 3}{2}, \frac{- 25}{4})$ $(\frac{-3}{2}, \frac{-25}{4})$

GRAPH
graph{x^2+3x-4 [-9.29, 10.71, -8, 2]}

Your graph should look like that.

How do you graph (x-1)(x+4)-y=0(x−1)(x+4)−y=0 (x-1)(x+4)-y=0?