How do you graph $y=-(x+3)^2 + 5$ ?

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How do you graph $y=-(x+3)^2 + 5$ ?

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Answer 1 · 2017-12-17T06:08:18

Vertex: $(-3,5)$

Y-intercept: $(0,-4)$

X-intercepts: $(-3+sqrt5,0),$ $(-3-sqrt5,0)$

Approximate x-intercepts: $x~~(-0.7639,0),$ $"and"$ $~~(-5.236,0)$

Explanation:

Graph:

$y=-(x+3)^2+5$ is a quadratic equation in vertex form:

$y=a(x-h)^2+k$ ,

where:

$a=-1$ , $h=-3$ , and $k=5$

The vertex, $(h,k)$ , is $(-3,5)$

To find the y-intercept, substitute $0$ for $x$ and solve for $y$ .

$y=-1(0+3)^2+5$

$y=-9+5$

$y=-4$

Y-intercept: $(0,-4)$

To find the x-intercepts, substitute $0$ for $y$ and solve for $x$ .

$-(x+3)^2+5=0$

$-(x+3)^2=-5$

Divide both sides by $-1$ .

$(x+3)^2=(-5)/(-1)$

Simplify.

$(x+3)^2=5$

Take the square root of both sides.

$x+3=+-sqrt5$

Subtract $3$ from both sides.

$x=-3+-sqrt5$

X-intercepts: $(-3+sqrt5,0),$ $(-3-sqrt5,0)$

$x~~(-0.7639,0),$ $"and"$ $~~(-5.236,0)$

Summary:

Vertex: $(-3,5)$

Y-intercept: $(0,-4)$

X-intercepts: $(-3+sqrt5,0),$ $(-3-sqrt5,0)$

Approximate x-intercepts: $x~~(-0.7639,0),$ $"and"$ $~~(-5.236,0)$

Plot the points and sketch a parabola through the points. Do not connect the dots.

graph{y=-(x+3)^2+5 [-10, 10, -5, 5]}

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How do you graph $y=-(x+3)^2 + 5$ ?

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How do you graph y=-(x+3)^2 + 5 y=-(x+3)^2 + 5?

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Explanation:

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How do you graph $y=-(x+3)^2 + 5$ ?