Question

If the equation of the tangent to the parabola `y=cx^2+d` is y=2x+3, how do you find the constants c and d and x=1?

Answer 1

If `y = cx^2+d`
which is the same as `y=cx^2+0*x+d`
then
`(dy)/(dx) = 2cx+0`
which we are told is
`=2x+3`
which (among other things)
implies `0=3`!!!!!!

Since this doesn't seem likely
I will assume that the intended parabolic equation was intended to be:
`y=cx^2+dx`
from which we get
`(dy)/(dx) = 2cx+d`
and given that this is
`=2x+3`
`rarr c=1 " and"`
`rarr d=3`

Your question ended with "...and x=1"
Again, I will make an assumption; namely that you wanted the value of the equation when `x=1`

Assuming all my assumptions are correct
the equation is
`y=x^2+3x`

and
`y_(x=1) = 1^2+3(1)` = 4#