MATH TIME!

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:

 

That part makes sense I just don't understand why it's a x-4 :/ instead of x+4

 

Thought abs was suppose to be positive inside!

 

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|

 

Interesting was trying to solve got everything except mixed up the sign and didn't know why T.T

 

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

 

Sweet . I'll have more for you soon! Hahaha

 

With your help @MarkFL I feel like I can obtain my 4.0 GPA

 

@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

 

Is this the expression?

 

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

 

I'm using my local server where I have a vB site running LaTeX powered by MathJax.

 

Snap that was awesome, yeah it's 9/4a+4

 

So I almost had it right I should have just timesed the fractions instead of everything.

 

Yeah, if you distribute the 1/4 then you do get (9/4)a + 4.

I actually prefer the first form I gave.

 

@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....