#help math pls help i forgot how to do this type of questions

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zenith idol
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i have graph (x-3)^2 +4
and the tangent to the graph at point A passes through the origin and I need to find the x-coordinate of point A give answer as surd

abstract spruceBOT
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zenith idol
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i did dy/dx and got 2x-6

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and turning point is (3,4) then did y=x(2x-6)

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and made (x-3)^2+4=2x^2-6x but I got x=-13 and x=1

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idk what i'm doing 😭

merry crystal
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The first step is to read the question and understand what it's asking for.

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And actually, this might be a good opportunity to post the question in its original text, i.e. a picture or screenshot.

zenith idol
merry crystal
zenith idol
merry crystal
merry crystal
zenith idol
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i get sqrt(13)

merry crystal
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The problem of understanding the problem can't be solved by "I do this instead".

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It's not about what you do, it's about how you think.

dull lance
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if you keep telling us random things we really cannot be useful

languid birch
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you already found the function that represents the gradient

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(the derivative)

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i'm sure you know what that does right

merry crystal
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And the conversation has already progressed way past where you're at.

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At this point we're just waiting on OP to, y'know, participate.

languid birch
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didn't notice

zenith idol
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i got answer i take derivative for gradient then

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equate line tangent 0,0

languid birch
languid birch
zenith idol
languid birch
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well, you are given the gradient function right? and a y-coordinate could just be the function which is y=(x-3)^2+4

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consider a point x=a

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and try and form a relationship between them, i guess

zenith idol
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they use y = mx+C method

languid birch
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well, i guess the form y-y_1 =m(x-x_1) would be more helpful\

zenith idol
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ok i got answer

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i get a quadratic

languid birch
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yes

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then you can solve for that quadratic and compare solutions

zenith idol
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thank you sir

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+close

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languid birch
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nw

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