Diagram shows the graph of y=8-2(x-a)^2.
find the value of a and b..
First look for the y-intercept: Set x = 0 that gives you y = ?
Then you wan´t to find the x intercepts. So set y = 0
Then 2(x-2)^2 must equal 8..
But you graph has a maximum of 8 when x = 1 seting x = 1 into the equation gives us:
$\displaystyle y = 8- 2(1-2)^2$
$\displaystyle Y = 8-2(-1)^2
y = 8-2 = 6 ???$
Either the graph is not entierly correct or the equation is not..
You have an equation $\displaystyle y = 8-2(x-a)^2$
Then you have a vertex at (1;8)
Then means when x = 1 then y = 8
Now if you enter those values into the equation that is: $\displaystyle 8 = 8-2(1-a)^2$
Now you have one equation and one unknown.
Then you solve for a.
$\displaystyle 8 = 8-2(1-a)^2$
$\displaystyle 8 = 8-2(1-a)^2$. notice that on the left side equals 8. then the right side must equal eight.
The only way the right side can equal 8 is when the expression $\displaystyle -2(1-a)^2$ = 0
that gives you 8 = 8.
For what values of a does the exspression equal 0???
There is only one value for a that is possible....
Once you done that you want to find b.
To find b. Enter x = 0 into your equation. like this:
$\displaystyle y = 8-2(0-a)^2$ by now you know what a is so whatever value comes out that is your b. also called the y intercept.
Can you take it from here?