1. ## Sketching graph of quadratic function

Diagram shows the graph of y=8-2(x-a)^2.

find the value of a and b..

2. ## Someting wrong with the picture?

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:
$y = 8- 2(1-2)^2$
$Y = 8-2(-1)^2
y = 8-2 = 6 ???$

Either the graph is not entierly correct or the equation is not..

3. Originally Posted by Henryt999
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:
$y = 8- 2(1-2)^2$
$Y = 8-2(-1)^2
y = 8-2 = 6 ???$

Either the graph is not entierly correct or the equation is not..
hello..sorry,i write a wrong question.but i have edit it .

4. ## OK no prob

You understood how to find b? Set x = 0
But first you must find a.
Do that by entering the cordinates for the vertex at (1;8)
Set x = 1; and y = 8 into the above equation..then solve for a.
Can you take it from here?

5. can u teach me step by step?

6. ## Sure.

You have an equation $y = 8-2(x-a)^2$
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: $8 = 8-2(1-a)^2$
Now you have one equation and one unknown.
Then you solve for a.
$8 = 8-2(1-a)^2$
$8 = 8-2(1-a)^2$. 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 $-2(1-a)^2$ = 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:
$y = 8-2(0-a)^2$ 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?

7. thanks a lot