I’m not totally sure about the less and greater signs because I’m not sure you can graph using those signs but either way their are your line equations
The equation in standard form is 2x^2 + 7x - 15=0. Factoring it gives you (2x-3)(x+5)= 0. That's the first one. The second one requires you to now your formula for the axis of symmetry which is x = -b/2a with a and b coming from your quadratic. Your a is -1 and your b is -2, so your axis of symmetry is
x= -(-2)/2(-1) which is x = 2/-2 which is x = -1. That -1 is the x coordinate of the vertex. You could plug that back into the equation and solve it for y, which is the easier way, or you could complete the square on the quadratic...let's plug in x to find y. -(-1)^2 - 2(-1)-1 = 0. So the vertex is (-1, 0). That's the first choice given. For the last one, since it is a negative quadratic it will be a mountain instead of a cup, meaning it doesn't open upwards, it opens downwards. Those quadratics will ALWAYS have a max value as opposed to a min value which occurs with an upwards opening parabola. This one is also the first choice because of the way the equation is written. There is no side to side movement (the lack of parenthesis tells us that) so the x coordinate for the vertex is 0. The -1 tells us that it has moved down from the origin 1 unit; hence the y coordinate is -1. The vertex is a max at (0, -1)
Answer:
The answer is 12x because 6x and 6x are like terms (same variable)
For part (a), you have


If

, then

.
If

, then

.
So,

For part (b), since the degrees of the numerator and denominator are the same, you first need to find the quotient and remainder upon division.

In the remainder term, the denominator

can't be factorized into linear components with real coefficients, since the discriminant is negative

. However, you can still factorized over the complex numbers, so a partial fraction decomposition in terms of complexes does exist.



Then you have


When

, you have



When

, you have



So, you could write

but that may or may not be considered acceptable by that webpage.