What is the minimum point of the graph of the equation y=2x^2+8x+9 ? plz help !!!
what they are asking for is the second coordinate of the vertex
the first coordinate of the vertex of \[y=ax^2+bx+c\] is always \[-\frac{b}{2a}\]
the second coordinate (what is you are looking for) is what you get when you replace \(x\) by \(-\frac{b}{2a}\)
@statellite73 i wasn't taught how to do it using that method
in your example \(a=2,b=8\) so first find \(-\frac{b}{2a}\)
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 \(-\frac{b}{2a}\)?
-2
yes it is
so now what is \[y=2(-2)^2+8\times (-2)+9\]?
Y=1
I used the calculator
yes it is 1 with or without a calculator
\[y=2(-2)^2+8\times (-2)+9\\ 2\times 4-16+9\\ 8-16+9\\ -8+9\\1\]
and that is your final answer
(-2,1)
btw this also means \[y=2x^2+8x+9=2(x+2)^2+1\]
Kk :)
yes, that is your "minimum point" or as we "vertex"
don't forget the magic formula \(\color{red}{-\frac{b}{2a}}\)
Alrighty THANK YOU so much !!! I really appreciated your help
yw
Wait... Quick question !
ok ..
Lol.. okay the "magic formula" can only be used when trying to find the minimum point?
it is the first coordinate of the vertex of any quadratic , i.e. anything that looks like \[y=ax^2+bx+c\]
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
for example if \[y=-3x^2+12x-4\]then \(a=-3,b=12\) and \(-\frac{b}{2a}=-\frac{12}{2\times (-3)}=2\)
since \[y=-3x^2+12x-4\] opens down, it would be the first coordinate of the highest point on the graph
Ahh okay I see THANKK YOU SO MUCHH !!! You're simply amazing 😝😝😝
@satellite73
you are quite welcome always nice to help with someone who actually participates !!
Wait another question.. this is my last one !!!!
@satellite73 do you have any tips on factoring ?
yes set it equal to zero and solve, then factor
you mean factoring a quadratic right? like \[x^2-x-6\] or sommat
Quadratic
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
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 \[(2x+1)(x-5)\]
Alrighty :) once again thanks a lot !!
once again, yw
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