Question

Which of the following could be the equation of the quadratic shown below?Explain your reasoning

(1) y=-3x² +8x-5
(2) y=4x2 - 6x +7
(3) y=-2x+12x+11
(4) y=x2-8x-2

Answer 1

Part 1) Option 3 could be the quadratic equation shown in the figure

Part 2) Option 4
`y\leq 11`

Explanation:

Part 1) we know that

The quadratic equation shown in the graph represent a vertical parabola open downward

The vertex represent a maximum

The coordinates of the vertex are positive

The y-intercept is positive

Has two real solutions (x-intercepts) one positive and one negative

In this problem, the options 2 and 4 represent a vertical parabola open upward (because the leading coefficient is positive)

so

Options 2 and 4 could not be the quadratic equation shown in the figure

Verify option 1 and 3

Option 1

`y=-3x^(2)+8x-5`

Find the y-intercept

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

so

For x=0

`y=-3(0)^(2)+8(0)-5`

`y=-5`

The y-intercept is negative

therefore

Option 1 could not be the quadratic equation shown in the figure

Option 3

`y=-2x^(2)+12x+11`

Verify the y-intercept

Find the y-intercept

For x=0

`y=-2(0)^(2)+12(0)+11`

`y=11`

The y-intercept is positive

Verify the vertex

Convert to vertex form

`y=-2x^(2)+12x+11`

Factor -2

`y=-2(x^(2)-6x)+11`

Complete the square

`y=-2(x^(2)-6x+9)+11+18`

`y=-2(x^(2)-6x+9)+29`

rewrite as perfect squares

`y=-2(x-3)^(2)+29`

The vertex is the point (3,29)

so

Both coordinates are positive

Verify the x-intercepts

Remember that the x-intercepts are the values of x when the vakue of y is equal to zero

For y=0

`-2(x-3)^(2)+29=0`

`2(x-3)^(2)=29`

`(x-3)^(2)=14.5`

square root bot sides

`x-3=(+/-)√(14.5)`

`x=3(+/-)√(14.5)`

Has two real solutions (x-intercepts) one positive and one negative

therefore

Option 3 could be the quadratic equation shown in the figure

Part 2) we know that

Using a graphing tool

Plot the points

The quadratic equation represent a vertical parabola open downward

The vertex is a maximum

so

The maximum value of y is equal to 11 (based in the table)

so

`y\leq 11`

see the attached figure to better understand the problem