Okay, suppose we are given a parabolic function in the form: y = a(x - h)^2 + k, where a ≠ 0 The vertex (for a parabola, this is the turning point) will be at (h,k). If a < 0 then the parabola will open downwards and the vertex will be at the maximum, but if 0 < a then the parabola opens upwards and the vertex will be at the minimum. The parabola we've been given may be written as: y = 1(x - 0)^2 + (-7) This means the vertex is at (0,-7) and the parabola opens upwards, so the minimum value is -7. Let's look at a graph of the function: The blue curve is the parabola, and the dots on the curve represent the 4 smallest values of the parabola when x is an integer. The horizontal lines represent the value of the parabola for those integral values of x, namely: y = -7 y = -6 y = -3 y = 2 Does that make sense?

Yes I remember this!!!!! so in other words I should go to my calculator for this type of question as well correct?

You shouldn't need a calculator for this, except perhaps if you want to graph the function to get an idea what the curve looks like and how it behaves.

Hey @MarkFL, I'm having some issues again with ABS I'm trying to understand why there is a minus instead of a plus |x-4| The distance between x and 4 is greater than 0 but less than 4. Is it because it's less than four? 0<|x-4|<4

We've got two cases to consider here...the first is: x - 4 ≥ 0 Recall that when |x| = x when x ≥ 0, and |x| = -x when x < 0. In this case, we can write: 0 < x - 4 < 4 Add through by 4: 4 < x < 8 The second case is: x - 4 < 0 In this case we may write: 0 < 4 - x < 4 Multiply through by -1: 0 > x - 4 > -4 Add through by 4: 4 > x > 0 And so our solution, in interval notation, is: (0, 4) U (4, 8) On the number line, this is:

The distance between x and 4 is the magnitude of the difference between x and 4. For a difference we use subtraction, and for a magnitude we use absolute value, thus the distance d between x and 4 on the number line is: d = |x - 4|

No, the actual absolute value is never negative, we always have: 0 ≤ |x| But the value of x itself can be any real number. If x is negative then: |x| = -x If x is non-negative, then: |x| = x For example, if we have: |-2| Then we see -2 < 0 and so: |-2| = -(-2) = 2

@MarkFL Okay my brain hurts I feel like I'm making a small error. Simplify the variable equation: 3/4(5a+2)-1/2(3a-5) Now I found the LCD to get rid of the fractions but I feel that's not what I was suppose to do... (I have the answer ). helppp meeee pleassseee <3

Nope, it's fractions on the outside of the brackets! Where do you get this images? I'll capture the questions lol

@MarkFL Having some issues with this one, I don't know why it looks really easy 16t^2+80t+4 t=2 I'm ending up going the easy way on this but should I be using quadratic equation? Ending up with 228....