Question

I'd like to know how i can tackle this question because its getting me confused:
" The curve y =1/a + bx passes through the point (1, -1) and its gradient at that point is 2. Find values of a and b

Answer 1

Task. The curve $y=1/a+bx$ passes through the point $(1,-1)$ and its gradient at that point is $2$. Find values of $a$ and $b$.

Solution. Probably instead of “gradient” there should be “derivative”. In this case the problem can be solved as follows.

Since the curve passes through the point $(-1,1)$, we have that

$y(-1)=1,$

that is

$1/a+b(-1)=1,$

$1/a=1+b,$

$a=\frac{1}{1+b}.$

Moreover,

$y^{\prime}(x)=(1/a+bx)^{\prime}=b.$

In particular,

$2=y^{\prime}(-1)=b,$

so

$b=2.$

Therefore

$a=\frac{1}{1+b}=\frac{1}{1+2}=\frac{1}{3}.$

Thus

$y(x)=3+2x.$

Answer. $a=1/3$, $b=2$.