Graph each quadratic equation!

Hi there,

I have solved quadratic equation but not sure this is correct or not please veiw it and let me know thanks!


Graph each quadratic equation. State the vertex and intercepts.

y = 1 - x^2


First find the x-coordinate of the vertex:
x = -b/2a = -(-1)/2(1) = 1/2

If x = 1/2, then
y = 1/2 - (1/2)^2
y = 1/2 -1/4
y = 2

So the vertex is (1/2, 2). If x = 0, then
y = 1 - 0^2
y = 1


The y-intercept is (0, 1). To find the x-intercepts, replace y by 0:

1 - x^2 = 0
-(x - 1)(x + 1)=0
x - 1 = 0 or x + 1 = 0
x = 1 or x = -1

First find the x-coordinate of the vertex:
x = -b/2a = -(-1)/2(1) = 1/2
If x = 1/2, then
y = 1/2 - (1/2)^2
y = 1/2 -1/4
y = 2

Click to expand...

???????????

If x = 1/2, then
y = 1/2 - (1/2)^2

Click to expand...

Should have read
If x = 1/2, then
y = 1 - (1/2)^2

The general equation is
y = ax^2+bx+c
Yours can be written
y = - x^2 + 0x + 1
so a = -1, b = 0, c = 1
-b/2a = -0/-2 = 0
y(0) = 1
That gives both the vertex and y intercept at P(0,1)

Your x intercepts are correct though.

Hi Gene,


how to solve this if x - 1/2 then

y = 1 - (1/2)^2

y = 1 - 1/4 = 4 or 1/4 ??? i forgot how to solve this please help me Thanks!

You have to do the common denominator thing.
1 = 4/4 so
1 - 1/4 =
4/4-1/4 =
(4-1)/4 =
3/4
But you do know that the whole thing is wrong 'cause x is NOT equal 1/2.

Thank you Gene,

Well, Let rework on this method and see if i get different x value or might you can show me your work will be great.

Talk to u soon!
I appreciate your response!

That was my work.
-b/2a = 0 not 1/2 so everything you did from that point on was wrong. The method right, but based on bad data and had one mistake,
If -b/2a HAD been 1/2 then
y(1/2) = 1/2 - (1/2)²
would have read
y(1/2) = 1 - (1/2)²
but since -b/2a was 0 the true vertex was at
y(0) = 1-0² = 1
Since that is the same equation as y(0) for the y intercept they are the same. Ok?