Question 1022113: Hi! I have two questions. Well...kind of 1.5. Both involve finding vertex, x intercepts, and standard form. The first problem is
1/4x^2-2x-12
To find the vertex I did -(-2)/2*(1/4( -> 2/1/2 -> 4. Then I plugged in 2 where x is -> 1/4(2)-2(2)-12 and got -16. This would then make the vertex (4,-16). I'm wondering how I would get all of this into standard form, which would then allow me to find the x intercepts. I guess the fractions are just what throw me off.
The second question is 3/5(x^2+8x-5). I got to 3/5(x+4)^2-21, but from there I'm stuck.
I appreciate any help!
Answer by solver91311(24713)   (Show Source):
You can put this solution on YOUR website!
Generally I am pretty hard-over on the one problem per post rule, but I'm also very forgiving when a student shows their work.
Problem 1: Your calculation of the coordinates of the vertex were spot on; the vertex is indeed located at the point ) .
Put yourself in the lotus position and repeat until you believe: "A fraction is simply a number and any whole number can be written as a fraction".
The standard form of a quadratic function is
Where  ,  , and  are numbers (which we accept and believe according to today's positive affirmation can just as well be fractions as otherwise)
If we are concerned with finding the  -intercepts of the function, then we need to find the two values of , if they exist, that cause the value of the function, namely  , to be zero. So set the function equal to zero and create the standard quadratic equation:
Now you actually have an embarrassment of riches available to you to solve this. You could simply factor it straight up, but the presence of the fraction makes this a little tricky, so I would opt for multiplying both sides of the equation by the denominator of the fraction to eliminate it, and THEN factoring -- thus:
(x\ -\ 12)\ =\ 0)
Leading to
or
 .
You could also use the quadratic formula which always works, even for the "factoring challenged".
\ \pm\ \sqrt{(-2)^2\ -\ 4\left(\frac{1}{4}\right)(-12)}}{2\left(\frac{1}{4}\right)})
or
Problem 2:
Missed it by THAT much. The fractional lead coefficient must also be applied to the -21 constant, thus:
^2\ -\ \frac{63}{5})
Which is the vertex form, ^2\ +\ k) , and the vertex is at ) .
All quadratics ) have the same intercepts regardless of the value of  , hence solve:
This one doesn't factor, so the only choice is the quadratic formula. However, please note a characteristic of this particular quadratic: The lead coefficient and the constant term have opposite signs. This guarantees that there will be two distinct real roots. That is because  must be positive whenever and have opposite signs.
\ \pm\ \sqrt{(8)^2\ -\ 4(1)(-5)}}{2(1)})
John
My calculator said it, I believe it, that settles it
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