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Mathematics 9 Online
OpenStudy (anonymous):

What is one of the zeros of the function shown in this graph?

OpenStudy (anonymous):

OpenStudy (anonymous):

zeros are the same as x-intercepts of the function. It is when y=0, so you have points (x,0) . look at your graph, since it is a quadratic equation you have two zeros. From the hint that I gave you above, can you tell me these points?

OpenStudy (anonymous):

(-4,0)?

OpenStudy (anonymous):

nope, the graph does not pass through that point try again

OpenStudy (anonymous):

(-1,0)?

OpenStudy (anonymous):

make sure that the points are 'INCLUDED' in the graph.

OpenStudy (anonymous):

(-3,0) and (1,0)?

OpenStudy (anonymous):

yesh

OpenStudy (anonymous):

ohh okay could you help me with one more?

OpenStudy (anonymous):

OpenStudy (anonymous):

i'll try if i still have some time left

OpenStudy (anonymous):

I attached the question above

OpenStudy (anonymous):

do you know how to calculate for the discriminant?

OpenStudy (anonymous):

yes b^2-4ac

OpenStudy (anonymous):

ok good. do you know how to apply it? if the discriminant is equal to zero, what does it mean? if it is less than zero? if it is greater than zero?

OpenStudy (anonymous):

I know how to plug in the numbers but I dont know the rest of it

OpenStudy (anonymous):

if the discriminant is equal to zero, it means that it has ONE zero/root/x-intercept/solution. if it is less than zero, it means that the root nature of the quadratic equation is NOT REAL (imaginary). if it is greater than zero, it means that there are TWO real roots/zeros/xints/solutions. so try calculating for the discriminant of that equation and tell me what is the value of the discriminant.

OpenStudy (anonymous):

\[x ^{2}+3x-7=0\] a=1, b=3, c=-7 \[b ^{2}-4ac\] \[3^{2}-4(1)(-7)=37\]

OpenStudy (anonymous):

is 37 greater than, equal, or lesser than zero? so what does it mean? (refer to the info that i gave you)

OpenStudy (anonymous):

its greater than zero so it has two real roots?

OpenStudy (anonymous):

absolutely right ^_^

OpenStudy (anonymous):

yay(: thank you(: could I ask you one more question

OpenStudy (anonymous):

lol okay

OpenStudy (anonymous):

OpenStudy (anonymous):

could you help me with this one?

OpenStudy (anonymous):

sorry i was doing something. gimme a sec. i'll look at it.

OpenStudy (anonymous):

Oh im sorry to bother you

OpenStudy (anonymous):

no its okay :) so we are looking for the maximum point of this quadratic equation. do you know how to find the vertex(h,k)?

OpenStudy (anonymous):

vertex formula?

OpenStudy (anonymous):

yes or change the standard form into vertex form?

OpenStudy (anonymous):

Im not good at changing formulas

OpenStudy (anonymous):

alright, let's just use the vertex formula. its easier \(\Large (h,k) = (\frac{-b}{2a},k) = (x, r(x))\) we are solving for k or r(x)

OpenStudy (anonymous):

\(r(x)=10x-0.025x^2\) \(a\)= -0.025, \(b\)=10 plug into the formula \(\Large \frac{-b}{2a}\) to solve for the number of toys (\(x\)), when the revenue is at the maximum. then after you find \(x\) or the number of toys, substitute it in the original equation \(r(x)=10x-0.025x^2\), to solve for the maximum revenue.

OpenStudy (anonymous):

i finished calculating it, what did you get?

OpenStudy (anonymous):

\[\frac{ -b }{ 2a }=\frac{ -10 }{ 2*0.025 }=200\] \[r(200)=10*(200)-0.025*(200)^{2}=1000\]

OpenStudy (anonymous):

also remember that question says that the revenue is 'measured in thousands of dollars', so multiply your answer, which is right btw, by $1000

OpenStudy (anonymous):

oops i meant -200

OpenStudy (anonymous):

so it would be -1000

OpenStudy (anonymous):

no, the first one is right because you substitute \(0.025\) instead of \(-0.025\), so you should end up having a POSITIVE answer

OpenStudy (anonymous):

i mean the answer should be positive

OpenStudy (anonymous):

Ohh okay so then it would be 1,000,000?

OpenStudy (anonymous):

yep

OpenStudy (anonymous):

dont forget the unit

OpenStudy (anonymous):

therefore, your answer should be $1,000,000

OpenStudy (anonymous):

Okay thank you(:

OpenStudy (anonymous):

no problem :)

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