please help me in answering this quadratic functions.....^_^?

Question

1. y=2x squared by 2 +12x+10
2. y=x squared by 2 +18x+6
3. y=4x squared by 2 +64x+251
4. y=(2x-1) squared by 2
5. y+1=(x+3) squared by 2

please show your solution

the instruction is: Express the following quadratic functions from the standard to the vertex form. Identify the vertex of the parabola and graph.

Answer 1

Well, I won't do all five *for* you, but I think I can get you started.

Let's start by making sure we're both on the same page. A quadratic function y = f(x) is written in "standard form" if it looks like y = ax² + bx + c, and written in "vertex form" if it looks like y = a(x - h)² + k. (Examples 1, 2, and 3, are in standard form; Example 4 will be in vertex form once you factor out the 2 from the parentheses, and Example 5 will be in vertex form once you subtract 1 from both sides.) If a quadratic function is in vertex form, you can easily write it in standard form by just "multiplying it out" and isolating y on the left side; if it's in standard form, you need to complete the square to get it in vertex form.

When the equation is written in vertex form, finding the vertex is easy -- it's coordinates are (h, k). If it's in standard form, the x-coordinate of the vertex will be -b/2a, and you can plug that value into the function to get the y-coordinate... but since we're being asked to put each one in vertex form anyway, we needn't do that extra work.

Let's do Example 1: y = 2x² + 12x + 10.

We'll start by isolating the x-terms on the right hand side, and to make it easier to complete the square, we'll eliminate the coefficient on the x²-term by dividing it into both sides. This gives us:

(y - 10)/2 = x² + 6x.

To complete the square, we halve, then square, the coefficient of the x-term, and add that result to both sides. Half of 6 is 3, and 3² is 9, so:

(y-10)/2 + 9 = x² + 6x + 9

The right side is now a perfect square trinomial, and we can write it as:

(y-10)/2 + 9 = (x + 3)²

Now we solve for y -- resisting the temptation to convert the (x+3)² back to x² + 6x + 9, mind you!

(y-10)/2 = (x+3)² - 9

y-10 = 2(x+3)² - 18

y = 2(x+3)² - 8

This is vertex form, and the coordinates of the vertex are (-3, -8). From this point, it should be easy to graphy by plotting a few more points, especially keeping in mind that the axis of symmetry will be the vertical line x = -3.

By the way, we can use the original equation y = 2x² + 12x + 10 to confirm the location of the vertex. The x-coordinate will be -12/(2*2) = -3, and the y-coordinate will then be y = f(-3) = 2(-3)² + 12(-3) + 10 = 18 - 36 + 10 = -8.

The other problems can be done in much the same way. Hope that helps!

[CORRECTED 8/8 9:45 am]

Answer 2

Vertex form is y = a(x - h)^2 + k

1.)
y = 2x^2 + 12x + 10
y = (2x^2 + 12x) + 10
y = 2(x^2 + 6x) + 10
y = 2(x^2 + 6x + 9 - 9) + 10
y = 2((x + 3)^2 - 9) + 10
y = 2(x + 3)^2 - 18 + 10
y = 2(x + 3)^2 - 8

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2.)
y = x^2 + 18x + 6
y = (x^2 + 18x) + 6
y = (x^2 + 18x + 81 - 81) + 6
y = ((x + 9)^2 - 81) + 6
y = (x + 9)^2 - 81 + 6
y = (x + 9)^2 - 75

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3.)
y = 4x^2 + 64x + 251
y = (4x^2 + 64x) + 251
y = 4(x^2 + 16x) + 251
y = 4(x^2 + 16x + 64 - 64) + 251
y = 4((x + 8)^2 - 64) + 251
y = 4(x + 8)^2 - 256 + 251
y = 4(x + 8)^2 - 5

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4.)
y = ((2x - 1)^2)/2
y = (1/2)(2x - 1)^2

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5.)
y + 1 = ((x + 3)^2)/2
y + 1 = (1/2)(x + 3)^2
y = (1/2)(x + 3)^2 - 1