How do you graph the parabola $f (x) = - {(x + 4)}^{2} + 9$ $f(x)= -(x+4)^2 + 9$ using vertex, intercepts and additional points?

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Answer 1 · 2018-07-16T06:20:43

Vertex $(- 4, 9)$ $(-4,9)$
y-intercept $(0, - 7)$ $(0,-7)$
x-intercepts $(- 1, 0); (- 7, 0)$ $(-1,0); (-7,0)$

Given -

$f (x) = - {(x + 4)}^{2} + 9$ $f(x)=-(x+4)^2+9$

$y = - {(x + 4)}^{2} + 9$ $y=-(x+4)^2+9$

Vertex $(- 4, 9)$ $(-4,9)$
At $x = 0; y = - {(0 + 4)}^{2} + 9 = - 16 + 9 = - 7$ $x=0; y=-(0+4)^2+9=-16+9=-7$

y-intercept $(0, - 7)$ $(0,-7)$

x intercepts
At $y = 0; - {(x + 4)}^{2} + 9 = 0$ $y=0; -(x+4)^2+9=0$

$- {(x + 4)}^{2} = - 9$ $-(x+4)^2=-9$

Multiplying both sides by $- 1$ $-1$

${(x + 4)}^{2} = 9$ $(x+4)^2=9$

$x + 4 = \pm \sqrt{9} = \pm 3$ $x+4=+-sqrt9=+-3$

$x = 3 - 4 = - 1$ $x=3-4=-1$

$x = - 3 - 4 = - 7$ $x=-3-4=-7$

x-intercepts $(- 1, 0); (- 7, 0)$ $(-1,0); (-7,0)$

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How do you graph the parabola f(x)= -(x+4)^2 + 9f(x)=−(x+4)2+9f(x)= -(x+4)^2 + 9 using vertex, intercepts and additional points?