How do you write 2x^2+12x-4 in vertex form?

2(x + 3)^2 - 22

Step-by-step explanation:

2x^2 + 12x - 4

Start by making your "a" value equal to 1 by factoring the 2 from the first 2 terms in the standard form equation.

2(x^2 + 6x) - 4

Complete the square by using the formula (b/2)^2. Identify your "b" value, which is 6. Now you can complete the square.

((6)/2)^2 = 9

After completing the square, add 9 inside the parentheses and subtract 9 outside the parentheses. Since the 9 inside the parentheses is also being multiplied by 2, multiply the subtracted 9 by 2 as well.

2(x^2 + 6x + 9) - 4 - 9(2)

Factor the terms inside the parentheses by using the product/sum factoring method. This is where you find two of the same terms that multiply to "c" (9) and add to "b" (6).

In this case, positive 3 multiplies to 9 and adds to 6, so we will use the factors (x + 3)(x + 3), which is the same as (x + 3)^2.

2(x + 3)^2 - 4 - 9(2)

To finish off the problem, combine the like terms outside of the parentheses by multiplying 9 times 2 first and then subtracting -4 by 9(2).

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no oblique asymptote

step-by-step explanation:

to find the oblique asymptote of f(x)

here numerator has degree 1 and denomintor degree 2

degree of numerator < degree of denominator

hence there is no oblique asymptote

when the degree of the denominator > that of the numerator, we saw that the value in the denominator got so much bigger, so quickly,   giving us a horizontal asymptote of the x-axis. or we can write

f(x) = a polynomial of x + a proper fraction.

in this case only we have oblique asymptotes but not when numerator has less degree.