Match the polar equations with the graphs labeled i vi

Asked by admin @ 22/11/2021 in Mathematics viewed by 261 People

Drag the tiles to the correct boxes to complete the pairs. Not all tiles will be used.

Match each graph with its corresponding equation.
NEED THIS ASAP PLS HELP

Answered by admin @ 22/11/2021

we know that

The equation of a vertical parabola in vertex form is equal to

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

where

(h,k) is the vertex

if $a>0$ ----> the parabola open upward (vertex is a minimum)

if $a$ ----> the parabola open downward (vertex is a maximum)

case A) $y=-x^{2}+5$

In this problem we have

the vertex is the point $(0,5)$

$a=-1$

therefore

the parabola open downward (vertex is a maximum)

The graph with its corresponding equation in the attached figure

case B) $y=\frac{3}{2} x^{2}-4x$

convert to vertex form

Factor the leading coefficient

$y=\frac{3}{2}(x^{2}-\frac{8}{3}x)$

Complete the square. Remember to balance the equation by adding the same constants to each side.

$y+\frac{8}{3}=\frac{3}{2}(x^{2}-\frac{8}{3}x+\frac{64}{36})$

Rewrite as perfect squares

$y+\frac{8}{3}=\frac{3}{2}(x-\frac{4}{3})^{2}$

$y=\frac{3}{2}(x-\frac{4}{3})^{2}-\frac{8}{3}$

In this problem we have

the vertex is the point $(\frac{4}{3},-\frac{8}{3})$

$a=\frac{3}{2}$

therefore

the parabola open upward (vertex is a minimum)

The graph with its corresponding equation in the attached figure

case C) $y=-x^{2}+2x+4$

convert to vertex form

Group terms that contain the same variable, and move the constant to the opposite side of the equation

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

Complete the square. Remember to balance the equation by adding the same constants to each side.

$y-4-1=-(x^{2}-2x+1)$

$y-5=-(x^{2}-2x+1)$

Rewrite as perfect squares

$y-5=-(x-1)^{2}$

$y=-(x-1)^{2}+5$

In this problem we have

the vertex is the point $(1,5)$

$a=-1$

the y-intercept is $4$

remember that the y-intercept is the value of y when the value of x is equal to zero

therefore

the parabola open downward (vertex is a maximum)

The graph with its corresponding equation is not in the attached figure

case D) $y=\frac{1}{2} x^{2}+2x+3$

convert to vertex form

Group terms that contain the same variable, and move the constant to the opposite side of the equation

$y-3=\frac{1}{2} x^{2}+2x$

Factor the leading coefficient

$y-3=\frac{1}{2}(x^{2}+4x)$

Complete the square. Remember to balance the equation by adding the same constants to each side.

$y-3+2=\frac{1}{2}(x^{2}+4x+4)$

$y-1=\frac{1}{2}(x^{2}+4x+4)$

Rewrite as perfect squares

$y-1=\frac{1}{2}(x+2)^{2}$

$y=\frac{1}{2}(x+2)^{2}+1$

In this problem we have

the vertex is the point $(-2,1)$

$a=\frac{1}{2}$

therefore

the parabola open upward (vertex is a minimum)

The graph with its corresponding equation in the attached figure

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

convert to vertex form

Group terms that contain the same variable, and move the constant to the opposite side of the equation

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

Factor the leading coefficient

$y-3=-2(x^{2}-2x)$

Complete the square. Remember to balance the equation by adding the same constants to each side.

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

$y-5=-2(x^{2}-2x+1)$

Rewrite as perfect squares

$y-5=-2(x-1)^{2}$

$y=-2(x-1)^{2}+5$

In this problem we have

the vertex is the point $(1,5)$

$a=-2$

the y-intercept is $3$

remember that the y-intercept is the value of y when the value of x is equal to zero

therefore

the parabola open downward (vertex is a maximum)

The graph with its corresponding equation in the attached figure

case F) $y=\frac{1}{2} x^{2}+3$

In this problem we have

the vertex is the point $(0,3)$

$a=\frac{1}{2}$

therefore

the parabola open upward (vertex is a minimum)

The graph with its corresponding equation is not in the attached figure

therefore

the answer in the attached figure

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