Could someone please translate these equations and explain the steps for doing it?
y = x² - 8x + 15
y = 3x² + 25x + 42
Thanks.
It's not magic. You just have to look.
x² - 8x + 15 =
Obvious part:
(x __ ___)(x __ ___)
Factors of 15
1,15
3,5
That's it.
Can one get 8 from these factors?
1+15 = 16 -- Nope.
1-14 = -14 -- Nope
3-5 = -2 -- Nope
3+5 = 8 -- Yes!!
(x __ 3)(x __ 5)
15 > 0, so 3 and 5 need the same sign.
-8 < 0, so both must be negative.
x² - 8x + 15 = (x - 3)(x - 5)
You do the other one.
I'm guessing you mean in vertex form, that is, $\displaystyle y=a(x-h)^2+k$
To do the we need to complete the square, to do this we take half of the linear coeffeint and square it.
$\displaystyle y=x^2-8x+15$
So the linear coeffeint is $\displaystyle -8$ half of it is $\displaystyle \frac{-8}{2}=-4$ and now we square it to get $\displaystyle (-4)^2=16$ This is what we need to complete the square.
We will now add zero to the euqation in the form of $\displaystyle 0=(16-16)$
so we get
$\displaystyle y=x^2-8x+15+0=x^2-8x+15+(16-16)$
Will will now rearrange the terms to get
$\displaystyle y=x^2-8x+16 +15-16$
Grouping the first three terms together and factoring gives
$\displaystyle y=[x^2-8x+16]-1=(x-4)^2-1$
So we get
$\displaystyle y=(x-4)^2-1$ We can now identify a, h, and k
a=1 h=4 and k=-1
So we know that the vertext is located at the point $\displaystyle (h,k) \to (4,-1)$ and the the parabola opens up because a is postitive.
I hope this helps and is what you wanted.
If so here is a hint for #2 factor like this
$\displaystyle y=ax^2+bx+c=a(x^2+\frac{b}{a}x)+c$
and try to adapt the method used above.
Multiplication is commutative, so (x - 3)(x - 5) = (x - 5)(x - 3)
$\displaystyle 3x^2+25x+42$
To factor this trinomial, first multiply leading coef. times constant: $\displaystyle 3\times42$
Result: 126
Find two factors of 126 that add up to the middle term's coef. of 25.
That'd be 18 and 7.
Replace the middle term with these two factors:
$\displaystyle 3x^2+18x+7x+42$
Factor by grouping: $\displaystyle (3x^2+18x)+(7x+42)$
$\displaystyle 3x(x+6)+7(x+6) = (3x+7)(x+6)$
Swapping the binomials will not have an effect. However, if you were to multiply $\displaystyle ax^2+bx+c=0$ by -1, then the parabola would open in the opposite direction. The vertex would change, but the x-intercepts would be the same.
Did you need to see $\displaystyle y=3x^2+25x+42$ in vertex form? You can determine the zeros from the factored version in my previous post. They are $\displaystyle \{-6,\frac{-7}{3}\}$
I'm having some trouble with this question too.
y = 20x² - 3x - 9
I get this:
= 20x² + 12x - 15x - 9
= (20x² + 12x) (-15x - 9)
= 2(10x + 12x) 3(-5x - 3)
(I don't know what to do next)
But the answer is supposed to be:
y = (4x - 3) (5x + 3)
What am I doing right/wrong?
Just make a catalogue.
20
1*20
2*10
4*5
9
1*9
3*3
-9 < 0, so opposite signs
We're looking for a -3
1*1 - 9*20 -- too big
1*20 - 9*1 = 11
1*3 - 9*3 = -26
2*1 - 10*20 -- too big
2*20 - 10*1 = 30
2*3 - 10*3 = -27
4*1 - 5*20 -- too big
4*20 - 5*1 = 75
4*3 - 5*3 = -3 -- Aha!!!!
(4x - 3)(5x + 3)
Be systematic.
Note: I started witht the most extreme values, just for the practice. I don't recommend that if you REALLY want the quickest path to the answer.
Sorry TKHunny, I'm a little confused by your method.
Using those numbers I can write (4x - 3) (5x + 3). Is that all I have to do?
And then when I'm finished do I do this to confirm my answer?We're looking for a -3
1*1 - 9*20 -- too big
1*20 - 9*1 = 11
1*3 - 9*3 = -26
2*1 - 10*20 -- too big
2*20 - 10*1 = 30
2*3 - 10*3 = -27
4*1 - 5*20 -- too big
4*20 - 5*1 = 75
4*3 - 5*3 = -3 -- Aha!!!!