can someone help me w this?

[gray rose]

i can do parts a and b but i don't know how to do part c

[bright prairie]

in c it says Q is the point where there is a horizontal tangent

and in a you have found f'(x)

so equate f'(x) to 0, you will get two x values with can be Q, the positive value will be Q

get the y value by computing that x value of q

since Q is on L, you can now put the x and y values into your answer for (b), solve for the only unkown now a

and you have your answer

[gray rose]

tyyy you're the realest

[gray rose]

but the issue i have is that f'(x) still has an a in it so i don't know how to find the solutions
i put it in the quad formula $\dfrac{2 \pm\sqrt{16-12a}}{3}$

[cursive knot]

runningbrook

[gray rose]

like idk if that's right or not but how would i go from there?

unless that's just the values and it's big long ugly and stupid and still has a in it idk 😭

[bright prairie]

yeah u can still solve with how i explained id think, but you will just have to keep x and y values in terms of a untill the end

[gray rose]

so u said that the positive value is q right? but like how do i know which is positive and negative if i don't know a?

[bright prairie]

u could take 2nd derivative

if that eliminates a

oh wait

what did u get for b

[gray rose]

for b i got y=ax+6

[wraith jolt]

f'(x) = 3x²+4x+a
P(0,6) -> f'(0) = a
y = ax+6
yes that's correct