# Math Help - Very Simple Differentiation

1. ## Very Simple Differentiation

The tangent to the curve y=ax²+bx at the point where x=1 has a gradient 1 and passes through point (4,5). Find a and b. Question is frustrating me .

Here's what I've got.

dy/dx = 2ax+b

(For x=1) 2a+b=1

Using curve and other given point: 5=16a+4b

Now when I put these into a sim eqation and solve it's wrong :/ I have no idea why either.

2. Originally Posted by alibond07
The tangent to the curve y=ax²+bx at the point where x=1 has a gradient 1 and passes through point (4,5). Find a and b. Question is frustrating me .

Here's what I've got.

dy/dx = 2ax+b

(For x=1) 2a+b=1

Using curve and other given point: 5=16a+4b

Now when I put these into a sim eqation and solve it's wrong :/ I have no idea why either.
Your working up to here is fine so not sure where you went wrong. Using your equations I arrived at the right answer so I'm guessing it's an arithmetic problem

Rearrange the first equation $b = 1-2a$

$5=16a+4(1-2a) = 8a+4 \: \: \therefore \: \: a = \frac{1}{8}$

$b = 1-\frac{2}{8} = \frac{3}{4}$

Verify using (4,5) and f(x)

$y = \frac{16}{8}+\frac{4 \times 3}{4} = 2+3 = 5$

3. Originally Posted by e^(i*pi)
Your working up to here is fine so not sure where you went wrong. Using your equations I arrived at the right answer so I'm guessing it's an arithmetic problem

Rearrange the first equation $b = 1-2a$

$5=16a+4(1-2a) = 8a+4 \: \: \therefore \: \: a = \frac{1}{8}$

$b = 1-\frac{2}{8} = \frac{3}{4}$

Verify using (4,5) and f(x)

$y = \frac{16}{8}+\frac{4 \times 3}{4} = 2+3 = 5$
Same as what I got : / , in the book they state it as, A= -1 B=-3. Maybe a mistake in the book? That dosen't seem to happen ever though : /

4. Originally Posted by alibond07
Same as what I got : / , in the book they state it as, A= -1 B=-3. Maybe a mistake in the book? That dosen't seem to happen ever though : /
Try subbing in those values:

$y = 16(-1)+(-3)(4) = -16-12 = -28$

Obviously $-28 \neq 5$ so the book is wrong