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Mathematics 9 Online
OpenStudy (bellezabrilla):

What is the minimum point of the graph of the equation y=2x^2+8x+9 ? plz help !!!

satellite73 (satellite73):

what they are asking for is the second coordinate of the vertex

satellite73 (satellite73):

the first coordinate of the vertex of \[y=ax^2+bx+c\] is always \[-\frac{b}{2a}\]

satellite73 (satellite73):

the second coordinate (what is you are looking for) is what you get when you replace \(x\) by \(-\frac{b}{2a}\)

OpenStudy (bellezabrilla):

@statellite73 i wasn't taught how to do it using that method

satellite73 (satellite73):

in your example \(a=2,b=8\) so first find \(-\frac{b}{2a}\)

satellite73 (satellite73):

ok i am sure you can also do it by "completing the square" but this method actually will complete the square for you quickly and easily so if \(a=2,b=8\) then what is \(-\frac{b}{2a}\)?

OpenStudy (bellezabrilla):

-2

satellite73 (satellite73):

yes it is

satellite73 (satellite73):

so now what is \[y=2(-2)^2+8\times (-2)+9\]?

OpenStudy (bellezabrilla):

Y=1

OpenStudy (bellezabrilla):

I used the calculator

satellite73 (satellite73):

yes it is 1 with or without a calculator

satellite73 (satellite73):

\[y=2(-2)^2+8\times (-2)+9\\ 2\times 4-16+9\\ 8-16+9\\ -8+9\\1\]

satellite73 (satellite73):

and that is your final answer

OpenStudy (bellezabrilla):

(-2,1)

satellite73 (satellite73):

btw this also means \[y=2x^2+8x+9=2(x+2)^2+1\]

OpenStudy (bellezabrilla):

Kk :)

satellite73 (satellite73):

yes, that is your "minimum point" or as we "vertex"

satellite73 (satellite73):

don't forget the magic formula \(\color{red}{-\frac{b}{2a}}\)

OpenStudy (bellezabrilla):

Alrighty THANK YOU so much !!! I really appreciated your help

satellite73 (satellite73):

yw

OpenStudy (bellezabrilla):

Wait... Quick question !

satellite73 (satellite73):

ok ..

OpenStudy (bellezabrilla):

Lol.. okay the "magic formula" can only be used when trying to find the minimum point?

satellite73 (satellite73):

it is the first coordinate of the vertex of any quadratic , i.e. anything that looks like \[y=ax^2+bx+c\]

satellite73 (satellite73):

if the parabola opens up, i.e if \(a>0\) then it gives the minimum point on the graph if the parabola opens down, ie. if \(a<0\) it gives the maximum point it is real easy to use

satellite73 (satellite73):

for example if \[y=-3x^2+12x-4\]then \(a=-3,b=12\) and \(-\frac{b}{2a}=-\frac{12}{2\times (-3)}=2\)

satellite73 (satellite73):

since \[y=-3x^2+12x-4\] opens down, it would be the first coordinate of the highest point on the graph

OpenStudy (bellezabrilla):

Ahh okay I see THANKK YOU SO MUCHH !!! You're simply amazing 😝😝😝

OpenStudy (bellezabrilla):

@satellite73

satellite73 (satellite73):

you are quite welcome always nice to help with someone who actually participates !!

OpenStudy (bellezabrilla):

Wait another question.. this is my last one !!!!

OpenStudy (bellezabrilla):

@satellite73 do you have any tips on factoring ?

satellite73 (satellite73):

yes set it equal to zero and solve, then factor

satellite73 (satellite73):

you mean factoring a quadratic right? like \[x^2-x-6\] or sommat

OpenStudy (bellezabrilla):

Quadratic

satellite73 (satellite73):

right so in my simple example \[x^2-x+6\] you need to find two numbers whose product is \(-6\) and that add to \(-1\) since \(-6\) is negative one has to be negative, the other positive in this case \(-3\) and \(2\) work because \(-3\times 2=-6\) and \(-3+2=-1\) so \[x^2-x-6=(x-3)(x+2)\] but that was a simple one

satellite73 (satellite73):

if it is harder, use the quadratic formula to find the zeros,then you know how to factor for example, if you find the zeros are \(-\frac{1}{2}\) and \(5\) then it factors as \[(2x+1)(x-5)\]

OpenStudy (bellezabrilla):

Alrighty :) once again thanks a lot !!

satellite73 (satellite73):

once again, yw

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