How do you find the vertex and the intercepts for $f (x) = {(x + 3)}^{2} + 1$ $f(x)=(x+3)^2+1$ ?

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How do you find the vertex and the intercepts for $f (x) = {(x + 3)}^{2} + 1$ $f(x)=(x+3)^2+1$ ?

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Answer 1 · 2018-08-06T10:06:44

$vertex = (- 3, 1), no x-intercepts$ $"vertex "=(-3,1)," no x-intercepts"$

Explanation:

$the equation of a parabola in vertex form$ $"the equation of a parabola in "color(blue)"vertex form"$ is.

$∙ x y = a {(x - h)}^{2} + k$ $•color(white)(x)y=a(x-h)^2+k$

$where (h, k) are the coordinates of the vertex and a$ $"where "(h,k)" are the coordinates of the vertex and a"$
$is a multiplier$ $"is a multiplier"$

$y = {(x + 3)}^{2} + 1 is in this form$ $y=(x+3)^2+1" is in this form"$

$vertex = (- 3, 1)$ $color(magenta)"vertex "=(-3,1)$

$let x=0, for y-intercept$ $"let x=0, for y-intercept"$

$y = 9 + 1 = 10 \leftarrow y-intercept$ $y=9+1=10larrcolor(red)"y-intercept"$

$let y = 0, for x-intercepts$ $"let y = 0, for x-intercepts"$

${(x + 3)}^{2} + 1 = 0$ $(x+3)^2+1=0$

${(x + 3)}^{2} = - 1$ $(x+3)^2=-1$

$this has no real solutions hence there are no$ $"this has no real solutions hence there are no"$
$x-intercepts$ $"x-intercepts"$
graph{(x+3)^2+1 [-20, 20, -10, 10]}

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How do you find the vertex and the intercepts for $f (x) = {(x + 3)}^{2} + 1$ $f(x)=(x+3)^2+1$ ?

1 Answer

Explanation:

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How do you find the vertex and the intercepts for f(x)=(x+3)2+1f(x)=(x+3)^2+1?

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Explanation:

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How do you find the vertex and the intercepts for $f (x) = {(x + 3)}^{2} + 1$ $f(x)=(x+3)^2+1$ ?