Solve the Quadratic Equations show work/explain pleaase?

Question

I need help finding the y-intercepts, and the vertexs.

1) find the vertex and the y-intercept.
y=2x²+4x+2

2) find the vertex and the y-intercept.
y=-3x²+12x-9

How do you find it? Show/work explain please.
3)
For the 1) I got 2(x²+2x+1) is y=2(x+1)²+1 for the vertex. Which is (1,-1) but its wrong why?

Answer 1

Question 1
y = 2(x² + 2x + 1)
y = 2(x² + 2x + 1 - 1 + 1)
y = 2(x² + 2x + 1)
y = 2(x + 1)²
Vertex is when x = - 1
Vertex at (- 1 , 0)
y intercept at y = 2

Question 2
y = - (3x² - 12x + 9)
y = (- 3 )(x² + 4x - 3)
y = (- 3) (x² + 4x + 4 - 4 - 3)
y = (- 3) (x² + 4x + 4 - 7)
y = (- 3) (x² + 4x + 4) + 21
y = (- 3) (x + 2)² + 21
Vertex at x = - 2
Vertex at (- 2 , 21)
y intercept at - 12 + 21 = 9

Answer 2

If you look at the Cartesian plane, you'll notice that the y intercept is also where x=0.  If your equation is y=ax²+bx+c, the y intercept is c by definition.

The vertex is a little more difficult.  A quadratic equation can be considered to be a perfect square plus a constant; y=a(x+i)²+j, which yields y=ax²+2aix+(ai²+j), and the vertex is at x=-i.  If you want to convert the form ax²+bx+c to the other form, i=b/2a (I'll let you work out the equation for j).

So the value of i for equation 1 would be (4/4) or 1, and the vertex is at x=-1.  The rest are for you to do.

Answer 3

x² - 6x = 216 do you already know the quadratic formula? in case you're looking it confusing to factor, you could constantly in basic terms use the quadratic formula. x² - 6x - 216 = 0 x = [ -b +/- sqrt(b^2 - 4ac) ] / (2a) x = [ 6 +/- sqrt(36 - 4(a million)(-216) ] / (2) x = [ 6 +/- sqrt(36 + 864) ] / (2) x = [ 6 +/- sqrt(900) ] / 2 x = [ 6 +/- 30 ] / 2 x = 3 +/- 15 for this reason, x = { 3 + 15, 3 - 15 } x = { 18, -12 } meaning this would have factor as (x - 18)(x + 12) = 0 {not that this would rely now which you have discovered the concepts}

Answer 4

1.
y=2x^2 + 4x + 2
y=(x+2)(x+1)

intercepts y at (-2,0) and (-1,0)

2.
y=-3x^2 +12x -9
y=-3(x+3)(x+1)

intercepts y at (-3,0) and (-1,0)