Defining a cubic equation from a point with known coordinates and slope, and a second point with only slope and x

Question

This is similar to this question, however the key difference that does not allow me to use that answer is that one of my known points is not x = 0.
I am required to find the equation of a cubic function, and I am only given one point [10,25] with slope 5 and a second point [45,f(x)] with slope of 4.
My attempts so far have dealt with the general form of a cubic: $$f(x)=ax^3+bx^2+cx+d\\f'(x)=3ax^2+2bx+c$$ Using this, and my known information I am able to get the following equations: $$25=10^3a+10^2b+10c+d\\5=3a10^2+20b+c\\4=3a45^2+90b+c$$ However, I have one less equation than I need in order to solve for the unknowns simultaneously.

Answer 1

In comments, you received good explanations for sure but at the same time you can solve the problem expecting that, at some time, you will get the value of $f(45)$.

So the equations are $$a\, 10^3 +b\, 10^2+c\, 10+d =25$$ $$3\,a \,10^2+2\,b \,10 +c =5$$ $$a \,45^3 +b \,45^2+c \,45+d =f(45)$$ $$3\,a\, 45^2+2\,b\, 45 +c =4$$

Using the method of your choice, this would give $$a=\frac{365-2 f(45)}{42875}\qquad b=\frac{33 f(45)-6145}{8575}$$ $$ c=\frac{28775-108 f(45)}{1715}\qquad d=\frac{5(20 f(45)-5463)}{343}$$ So, you can write a small program (or use Excel) which, for a given value of $f(45)$ will output the coefficients $a,b,c,d$ and will be able to compute $f(x)$ for any other $x$.

It would be $$f(x)=\frac{-2 x^3+165 x^2-2700 x+12500}{42875}f(45)+\frac{73 x^3-6145 x^2+143875 x-682875}{8575}$$