Which coefficien in the function y=-2x^2+5x-4 tells us that the parabola opens downward? @satellite73
the "leading coefficient' which is the number in front of the \(x^2\) term
in this case it is \(-2\) and since \(-2\) is negative, the parabola opens down
WHAT ABOUT THIS ONE ? What are the x-intercepts for the equation: y = x2 + 3x − 4
\[y=x^2+3x-4=(x+4)(x-1)\] set \(x+4=0\) and \(x-1=0\) just like before you get \(x=-4\) or \(x=1\)
\(x\) intercept is a synonym for the "zeros" i.e. set it equal to zero and solve, like last night
-3,4?
no no \(-4,1\)
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?
do you have a choice that looks like \[y=-(x-1.5)^2+4.25\]?
no
or \(y=-x^2+3x+2\) ?
got that choice?
if not let me see the choices, but i bet you have \[y=-2x^2+3x+2\]
What is the vertex of the function y=x^2-4yx+3 help me with this pleassseeee
what is that \(y\) doing there? a typo?
\[y=x^2-4x+3\] is that it?
in this case \(a=1, b=-4\) and \(-\frac{b}{2a}=-\frac{-4}{2}=2\) first coordinate of the vertex is \(2\)
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 \((2,-1)\)
you got that?
yeah i did thanks C:
gotta do better than 70 tonight right?
yup :d my last module so happy
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