# Differentiation questions (1 Viewer)

#### smithjohn

##### New Member
a) Differentiate y=x^2+bx+c, and hence find b and c, given that the parabola is tangent to the x-axis at the point (5,0)

b) Differentiate y=x^2+bx+c, and hence find b and c, when x =3 the gradient is 5, and x=2 is a zero.

#### Drdusk

##### π
Moderator
a) Differentiate y=x^2+bx+c, and hence find b and c, given that the parabola is tangent to the x-axis at the point (5,0)

b) Differentiate y=x^2+bx+c, and hence find b and c, when x =3 the gradient is 5, and x=2 is a zero.
$\bg_white \text{a)}$

$\bg_white y' = 2x + b$

$\bg_white \text{Since it's tangent to the x axis at 5,0 it means the gradient of the parabola at 5,0 is 0 because the gradient of the x axis is 0}$

$\bg_white \therefore 0 = 2(5) + b \Rightarrow b = -10$

$\bg_white \text{Figure out c easily by just subbing the point (5,0) in the original equation of the parabola.}$

$\bg_white \text{Part b is done in basically the same way as part a)}$

#### TheShy

##### Member
How can you be sure that the gradient at the x-axis is always 0? Like if it cuts the x axis twice, then wouldn't it have a non-zero gradient?

#### Drdusk

##### π
Moderator
How can you be sure that the gradient at the x-axis is always 0? Like if it cuts the x axis twice, then wouldn't it have a non-zero gradient?
Because the parabola is a tangent at that point. A parabola can only ever be a tangent to the x axis if it has it's vertex at a point on the x axis. It can never touch it twice in that case because it's an increasing function only.

#### TheShy

##### Member
Because the parabola is a tangent at that point. A parabola can only ever be a tangent to the x axis if it has it's vertex at a point on the x axis. It can never touch it twice in that case because it's an increasing function only.
Ah mb, didn't read that the parabola was a tangent to the x-axis

#### smithjohn

##### New Member
$\bg_white \text{a)}$

$\bg_white y' = 2x + b$

$\bg_white \text{Since it's tangent to the x axis at 5,0 it means the gradient of the parabola at 5,0 is 0 because the gradient of the x axis is 0}$

$\bg_white \therefore 0 = 2(5) + b \Rightarrow b = -10$

$\bg_white \text{Figure out c easily by just subbing the point (5,0) in the original equation of the parabola.}$

$\bg_white \text{Part b is done in basically the same way as part a)}$
so for part b)

y'(x) = 2x+b

when x = 3, the gradient is 5

So

5=2(3)+b
5=6 +b
-1=b

Sub x=3 and b = -1 into y=x^2+bx+c
y=(3)^2+(-1)(3)+c
y=6+c
Then the second part

y'(x) = 2x+b

when x = 2, the gradient is 0

So

0=2(2)+b
0=4 +b
-4=b

Sub x=2 and b = -4into y=x^2+bx+c
y=(2)^2+(-4)(2)+c
y=-4+c

But if you equate both equation the c cancels out, what am I doing wrong?

#### TheShy

##### Member
Im pretty sure when it says that x=2 is a zero, it means that its an intercept, such that when x=2, y=0.

#### smithjohn

##### New Member
Im pretty sure when it says that x=2 is a zero, it means that its an intercept, such that when x=2, y=0.
Thanks