How do you graph the inequality $y \leq - x^{2} + 5 x + 6$ $y<=-x^2+5x+6$ ?

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How do you graph the inequality $y \leq - x^{2} + 5 x + 6$ $y<=-x^2+5x+6$ ?

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Answer 1 · 2017-10-12T18:04:58

it's a parabola

Explanation:

you can find the x-intercepts using the quadratic equation
$- b \pm \frac{\sqrt{{(b)}^{2} - 4 \cdot a \cdot c}}{2 \cdot a}$ $-b+-sqrt((b)^2-4*a*c)/(2*a)$
then find the vertex by using $\frac{- b}{2 \cdot a}$ $(-b)/(2*a$ and $f (\frac{- b}{2 \cdot a})$ $f((-b)/(2*a))$
Then to solve for the inequality, use the coordinates (0,0) for the x and y points, with your equation
$y \leq - x^{2} + 5 x + 6$ $y ≤ −x^2+5x+6$
you get $0 \leq 6$ $0 ≤ 6$
since this is true, the area that contains the coordinate (0,0) is the correct area

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How do you graph the inequality $y \leq - x^{2} + 5 x + 6$ $y<=-x^2+5x+6$ ?

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How do you graph the inequality y<=-x^2+5x+6y≤−x2+5x+6y<=-x^2+5x+6?

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Explanation:

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How do you graph the inequality $y \leq - x^{2} + 5 x + 6$ $y<=-x^2+5x+6$ ?