Answer:
Part 1) Option 3 could be the quadratic equation shown in the figure
Part 2) Option 4 ![y\leq 11](https://tex.z-dn.net/?f=y%5Cleq%2011)
Step-by-step 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
<u><em>Verify option 1 and 3</em></u>
Option 1
![y=-3x^{2}+8x-5](https://tex.z-dn.net/?f=y%3D-3x%5E%7B2%7D%2B8x-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](https://tex.z-dn.net/?f=y%3D-3%280%29%5E%7B2%7D%2B8%280%29-5)
![y=-5](https://tex.z-dn.net/?f=y%3D-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](https://tex.z-dn.net/?f=y%3D-2x%5E%7B2%7D%2B12x%2B11)
<em>Verify the y-intercept</em>
Find the y-intercept
For x=0
![y=-2(0)^{2}+12(0)+11](https://tex.z-dn.net/?f=y%3D-2%280%29%5E%7B2%7D%2B12%280%29%2B11)
![y=11](https://tex.z-dn.net/?f=y%3D11)
The y-intercept is positive
<em>Verify the vertex</em>
Convert to vertex form
![y=-2x^{2}+12x+11](https://tex.z-dn.net/?f=y%3D-2x%5E%7B2%7D%2B12x%2B11)
Factor -2
![y=-2(x^{2}-6x)+11](https://tex.z-dn.net/?f=y%3D-2%28x%5E%7B2%7D-6x%29%2B11)
Complete the square
![y=-2(x^{2}-6x+9)+11+18](https://tex.z-dn.net/?f=y%3D-2%28x%5E%7B2%7D-6x%2B9%29%2B11%2B18)
![y=-2(x^{2}-6x+9)+29](https://tex.z-dn.net/?f=y%3D-2%28x%5E%7B2%7D-6x%2B9%29%2B29)
rewrite as perfect squares
![y=-2(x-3)^{2}+29](https://tex.z-dn.net/?f=y%3D-2%28x-3%29%5E%7B2%7D%2B29)
The vertex is the point (3,29)
so
Both coordinates are positive
<em>Verify the x-intercepts</em>
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](https://tex.z-dn.net/?f=-2%28x-3%29%5E%7B2%7D%2B29%3D0)
![2(x-3)^{2}=29](https://tex.z-dn.net/?f=2%28x-3%29%5E%7B2%7D%3D29)
![(x-3)^{2}=14.5](https://tex.z-dn.net/?f=%28x-3%29%5E%7B2%7D%3D14.5)
square root bot sides
![x-3=(+/-)\sqrt{14.5}](https://tex.z-dn.net/?f=x-3%3D%28%2B%2F-%29%5Csqrt%7B14.5%7D)
![x=3(+/-)\sqrt{14.5}](https://tex.z-dn.net/?f=x%3D3%28%2B%2F-%29%5Csqrt%7B14.5%7D)
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](https://tex.z-dn.net/?f=y%5Cleq%2011)
see the attached figure to better understand the problem