A system of equations is graphed on the coordinate plane.
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The parabola will show the vertex in the format: y-k = (x-h)^2, where the vertex point
lies at (h, k).
let's first put it in "y =" standard format:
Since we cannot get a perfect square out of this, we complete the square: a=1, b=2, c=3
(b/2)^2 = (2/2)^2 = 1, so
So there's +2 leftover, since 3-1=2; so:
Now we'll subtract the 2 from both sides to show our vertex:
where our vertex (h, k) is at (-1, 2)
lies at (h, k).
let's first put it in "y =" standard format:
Since we cannot get a perfect square out of this, we complete the square: a=1, b=2, c=3
(b/2)^2 = (2/2)^2 = 1, so
So there's +2 leftover, since 3-1=2; so:
Now we'll subtract the 2 from both sides to show our vertex:
where our vertex (h, k) is at (-1, 2)
Answer:
or
Step-by-step explanation:
We use casework on when and when
.
For the first case, , we add 9 to both sides to get
.
Dividing both sides by 3 gives
For the second case, , we add 9 to both sides to get
.
Dividing both sides by 3 gives
.
Checking both cases, we plug in and
.
For the first case, we have , which satisfies the equation.
For the second case, we have , which also satisfies the equation.
This gives us two solutions to the equation; and
.