Question

Could someone help me find the vertex?

Answer 1

hmmm can you solve for "y"?

Answer 2

umm..plug in (0,0)?

Answer 3

I think you should find the it's zeros

Answer 4

Do what?

Answer 5

Like plug in 0 for x and go from there?

Answer 6

hmm well.. fistly solve for "y"

Answer 7

I am not good at math, but I think you should set this equation equal to zero and solve for x.

Answer 8

I don't know how to though lol, what do I need to do?

Answer 9

hmmm ok so how would you solve for "y" if it say were \(\bf 2y = -2x^2-7x+4 \quad ? \)

Answer 10

I would..divide by 2?

Answer 11

yeap

Answer 12

so now y = -x^2 -7/2x + 2

Answer 13

\(\bf 2y\le-2x^2-7x+4\implies y\le -\cfrac{2x^2}{2}-\cfrac{7}{2}+\cfrac{4}{2}\implies y\le -x^2-\cfrac{7}{2}+2\)

Answer 14

what happened to the x in 7/2?

Answer 15

nothing

Answer 16

hhmmm ohh yea shoot,... what happened? well, it got run over by a typo =)

Answer 17

\(\bf 2y\le-2x^2-7x+4\implies y\le -\cfrac{2x^2}{2}-\cfrac{7}{2}x+\cfrac{4}{2}\implies y\le -x^2-\cfrac{7}{2}x+2\)

Answer 18

so as you can see, is pretty much like you'd do with a linear equation more or less

so you have a quadratic equation, to solve quadratics, you'd set y = 0, so \(\bf 2y\le-2x^2-7x+4\implies y\le -\cfrac{2x^2}{2}-\cfrac{7}{2}x+\cfrac{4}{2}\\ \quad \\
y\le -x^2-\cfrac{7}{2}x+2\implies 0\le -x^2-\cfrac{7}{2}x+2\)

Answer 19

Okay I get that, so now what do we do?

Answer 20

to find the vertex in a quadratic of type \(\bf y = ax^2+bx+c\) you'd get it at the coordinates given by \(\bf \left(-\cfrac{b}{2a}\quad ,\quad c-\cfrac{b^2}{4a}\right)\)

Answer 21

"a" "b" and "c" being of course, the coefficients in the quadratic

Answer 22

so -(-7/2/2(-1), 2 - (-7/2)^2/4(-1))?

Answer 23

yeap

Answer 24

so I got -7/4, -17/16

Answer 25

hmmm

Answer 26

one sec

Answer 27

\(\bf \begin{array}{llll}
&a=-1&b=-\frac{7}{2}&c=2\\
0\le &-x^2&-\cfrac{7}{2}x&+2
\end{array}\\ \quad \\

\left(-\cfrac{b}{2a}\quad ,\quad c-\cfrac{b^2}{4a}\right)
\implies \left(-\cfrac{-\frac{7}{2}}{2(-1)}\quad ,\quad 2-\cfrac{\left(-\frac{7}{2}\right)^2}{4(-1)}\right)\\ \quad \\
------------------------\\
-\cfrac{-\frac{7}{2}}{2(-1)}\implies -\left(-\cfrac{7}{2}\cdot \cfrac{1}{-2}\right)\implies \color{blue}{-\cfrac{7}{4}}\\ \quad \\
2-\cfrac{\left(-\frac{7}{2}\right)^2}{4(-1)}\implies 2-\left(-\cfrac{49}{4}\cdot \cfrac{1}{-4}\right)\implies 2-\left(\cfrac{49}{-16}\right)\implies \color{blue}{\cfrac{32}{16}+\cfrac{49}{16}}\)

Answer 28

hmm.... anyhow.... I have ... a small... minus there... lemme fix that \((\bf 2-\cfrac{\left(-\frac{7}{2}\right)^2}{4(-1)}\implies 2-\left(\cfrac{49}{4}\cdot \cfrac{1}{-4}\right)\implies 2-\left(\cfrac{49}{-16}\right)\)

Answer 29

where did you get the 1/-4?

Answer 30

\(\bf \cfrac{\quad \frac{a}{b}\quad }{\frac{c}{d}}\implies \cfrac{a}{b}\cdot \cfrac{d}{c}\)

Answer 31

I dont' really understand how you got your answer for the second one

Answer 32

hmm.... lemme rewrite it some more

Answer 33

\(\bf 2-\cfrac{\left(-\frac{7}{2}\right)^2}{4(-1)}\implies
\cfrac{-\frac{7}{2}\cdot -\frac{7}{2}}{\frac{-4}{1}}\implies \cfrac{\frac{49}{4}}{\frac{-4}{1}}
\\ \quad \\
2-\left(\cfrac{49}{4}\cdot \cfrac{1}{-4}\right)\implies
2-\left(\cfrac{49}{-16}\right)\implies \cfrac{32}{16}+\cfrac{49}{16}\)

Answer 34

Okay I get that so far...how did you get 32/16 + 49/16? where did that come from?

Answer 35

I simply made the "2" the same fractional type as the other for the sake of simplicity in the addition

how many sixteenths in 2? well, 2 * 16 = 32, so 32

thus \(\bf \cfrac{32}{16}\Longleftrightarrow 2\)

Answer 36

I could have added just the same using the LCD

Answer 37

but we subtracted and not multiplied?

Answer 38

\(\bf a-b\implies a-(+b)\implies a-b\\ \quad \\
a+b\implies a-(-b)\implies a+b\)

Answer 39

ohh okay