Ask your own question, for FREE!
Mathematics 17 Online
OpenStudy (anonymous):

Which coefficien in the function y=-2x^2+5x-4 tells us that the parabola opens downward? @satellite73

OpenStudy (anonymous):

the "leading coefficient' which is the number in front of the \(x^2\) term

OpenStudy (anonymous):

in this case it is \(-2\) and since \(-2\) is negative, the parabola opens down

OpenStudy (anonymous):

WHAT ABOUT THIS ONE ? What are the x-intercepts for the equation: y = x2 + 3x − 4

OpenStudy (anonymous):

\[y=x^2+3x-4=(x+4)(x-1)\] set \(x+4=0\) and \(x-1=0\) just like before you get \(x=-4\) or \(x=1\)

OpenStudy (anonymous):

\(x\) intercept is a synonym for the "zeros" i.e. set it equal to zero and solve, like last night

OpenStudy (anonymous):

-3,4?

OpenStudy (anonymous):

no no \(-4,1\)

OpenStudy (anonymous):

oh okay got it help me with this last one please Which quadratic equation has a vertex of (1.5, 4.25) and opens downward?

OpenStudy (anonymous):

do you have a choice that looks like \[y=-(x-1.5)^2+4.25\]?

OpenStudy (anonymous):

no

OpenStudy (anonymous):

or \(y=-x^2+3x+2\) ?

OpenStudy (anonymous):

got that choice?

OpenStudy (anonymous):

if not let me see the choices, but i bet you have \[y=-2x^2+3x+2\]

OpenStudy (anonymous):

What is the vertex of the function y=x^2-4yx+3 help me with this pleassseeee

OpenStudy (anonymous):

what is that \(y\) doing there? a typo?

OpenStudy (anonymous):

\[y=x^2-4x+3\] is that it?

OpenStudy (anonymous):

in this case \(a=1, b=-4\) and \(-\frac{b}{2a}=-\frac{-4}{2}=2\) first coordinate of the vertex is \(2\)

OpenStudy (anonymous):

second coordinate is what you get when you replace \(x\) by \(2\) you should get \[2^2-4\times 2+3=4-8+3=-1\] so vertex is \((2,-1)\)

OpenStudy (anonymous):

you got that?

OpenStudy (anonymous):

yeah i did thanks C:

OpenStudy (anonymous):

gotta do better than 70 tonight right?

OpenStudy (anonymous):

yup :d my last module so happy

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!