Question

Determine the y-intercept of a line with a slope of -2 that is a tangent to the curve y=12x^2+4x+5

Answer 1

you did post this earlier -- i just reply as thought you got it; apparently you didnt

we know that \( y=12x^2+4x+5 \) right?

first, find the derivative of it:
\[ \frac{d}{dx}\left(12x^2+4x+5\right) \]
\[ =\frac{d}{dx}\left(12x^2\right)+\frac{d}{dx}\left(4x\right)+\frac{d}{dx}\left(5\right) \]
\[ =12\frac{d}{dx}\left(x^2\right)+4\frac{d}{dx}\left(x\right)+0 \]
\[ =12\cdot \:2x^{2-1}+4\cdot 1+0 \]
\[ =24x+4+0 \]
thus, \( \frac{dy}{dx}=24x+4 \)
knowing that we can now write that:
\[ -2=24x+4 \ \ \Rightarrow \ \ -6= 24x \ \ \Rightarrow \ \ x=-\frac{1}{4} \]
now that we know what x is, we can plug that in our main formula:
\[ y=12x^2+4x+5 \ \ \Rightarrow \ \ y=12\left(\frac{1}{4}\right)^2+4\left(\frac{1}{4}\right)+5 \]
\[ y=12\left(\frac{1}{4}\right)^2+4\left(\frac{1}{4}\right)+5 \ \ \Rightarrow \ \ y=\frac{3}{4}+6 \]
\[ y=\frac{6\cdot \:4}{4}+\frac{3}{4} \Rightarrow y=\frac{27}{4} \]
and we can see that we have got this coordinate: \( ( y=-\frac{1}{4} ,\frac{27}{4} ) \) and the slope which is \( m=-2 \)
lets plug all that information into the slope formula:
\[ -2 = \frac{ (y-\frac{27}{4}) }{(x-(-\frac{1}{4}))} \]
\[ -2(x-(-\frac{1}{4})) = (y-\frac{27}{4}) \]
\[-2x-\frac{1}{2}= (y-\frac{27}{4})\]
\[ -2x=y-\frac{27}{4}+\frac{1}{2} \ \ \ \ \Rightarrow -2x=y-\frac{25}{4} \]
\[ x= \frac{y}{-2}-\frac{\frac{25}{4}}{-2} \]
\[ x=-\frac{4y-25}{8}\ \ \ \ \Rightarrow \ \ \ \ y=-\frac{8x-25}{4}\]
that y intercept should be at \( x=0 \), thus we write:
\[ y=-\frac{\cancel{8(0)}-25}{4} \]
\[ y=\frac{25}{4} \]
all that said, the y intercept is going to be at \( (0,\frac{25}{4} ) \)

Answer 2

you could have posted a new one for that, but since we are here, as discussed in dms, here is the answer to this one as well -- lazy to latex it, so I just handwrote it haha

Answer 3

Thank you so much!! this is very helpful. I can use this example on other questions.