Answer :
Answer:
[tex]\frac{145}{24}\approx 6[/tex]***
Answers may vary depending on the person reading the graph.
It is hard to tell what they think crosses nicely.
For example, I decided that the graph crossed nicely at the following points:
[tex](0,3)[/tex]
[tex](2,8)[/tex]
[tex](6,5)[/tex]
Step-by-step explanation:
I see that the graph crosses at [tex](0,3)[/tex],[tex](2,8)[/tex], and [tex](6,5)[/tex].
The equation [tex]y=ax^2+bx+c[/tex] tells us the [tex]y[/tex]-intercept is [tex]c[/tex] since when [tex]x=0[/tex] we have [tex]y=a(0)^2+b(0)+c=0+0+c=c[/tex].
The [tex]y[/tex]-intercept for our graph is [tex]3[/tex]. Therefore, [tex]c=3[/tex].
So far we have the equation:
[tex]y=ax^2+bx+3[/tex].
Let's enter the other points creating a system of linear equations to solve:
[tex]8=a(2)^2+b(2)+3[/tex]
[tex]5=a(6)^2+b(6)+3[/tex]
Let's simplify:
[tex]8=4a+2b+3[/tex]
[tex]5=36a+6b+3[/tex]
Let's subtract 3 on both sides:
[tex]5=4a+2b[/tex]
[tex]2=36a+6b[/tex]
I choose to solve the system by elimination.
Let's multiply the top equation by -3:
[tex]-15=-12a-6b[/tex]
[tex]2=36a+6b[/tex]
Now adding the equations results in:
[tex]-13=24a[/tex]
Divide both sides by 24:
[tex]\frac{-13}{24}=a[/tex]
Now using one of the equations we can find [tex]b[/tex]:
[tex]5=4a+2b[/tex] with [tex]a=\frac{-13}{24}[/tex]
[tex]5=4\frac{-13}{24}+2b[/tex]
[tex]5=\frac{-13}{6}+2b[/tex]
Add 13/6 on both sides:
[tex]5+\frac{13}{6}=2b[/tex]
[tex]\frac{30+13}{6}=2b[/tex]
[tex]\frac{43}{6}=2b[/tex]
Divide both sides by 2:
[tex]\frac{43}{2(6)}=b[/tex]
[tex]\frac{43}{12}=b[/tex]
So we have the equation:
[tex]y=\frac{-13}{24}x^2+\frac{43}{12}+3[/tex]
Let's evaluate [tex]a+b+c[/tex] now:
[tex]\frac{-13}{24}+\frac{43}{12}+3[/tex]
[tex]\frac{-13+86+72}{24}[/tex]
[tex]\frac{73+72}{24}[/tex]
[tex]\frac{145}{24}[/tex]