# Thread: A couple of easy ones

1. ## A couple of easy ones

Hi All,

Am having a real brain fade tonight, can you help me with the following:

Differentiate y = x² + bx + c, and hence find b and c if:

e) the parabola is tangent to 3x + y – 5 = 0 at the point T(3,-1)

f) the line 3x + y – 5 = 0 is a normal at the point T(3,-1)

Thanks heaps

2. Originally Posted by Joel
Hi All,

Am having a real brain fade tonight, can you help me with the following:

Differentiate y = x² + bx + c, and hence find b and c if:

e) the parabola is tangent to 3x + y – 5 = 0 at the point T(3,-1)

f) the line 3x + y – 5 = 0 is a normal at the point T(3,-1)

Thanks heaps
$
y=x^2+bx+c
$

$\frac{dy}{dx}=2x+b$ , this is the gradient of the tangent .

$2x+b=-3$ ( Its the gradient of the line $y=-3x+5$)

At T(3,-1) , $2(3)+b=-3$ .. Solve for b .

Now substitute the value of b , x and y into the original equation to look for c .

For f , make use of $m_1m_2=-1$

3. Hi.

Originally Posted by Joel
Hi All,

Am having a real brain fade tonight, can you help me with the following:

Differentiate y = x² + bx + c, and hence find b and c if:
I guess you already know that

y' = 2x + b

Instead of y and y' I better use

$f(x) = x^2+bx+c$

$f'(x) = 2x+b$

Originally Posted by Joel
e) the parabola is tangent to 3x + y – 5 = 0 at the point T(3,-1)

First consider 3x+y-5 = 0, then you get (adding +5 to both sides, -3x to both sides)

y = -3x+5

=> slope m = -3

Alright, the parabola f(x) goes through the point T(3,-1)

=> $f(3) = 3^2 + b*3 +c = 1$ (equation 1)

and the parabola has the slope -3 in T(3,-1)

Thus f'(3) = 2*3+b = -3 (equation 2)

You have two equations

$3^2 + b*3 +c = 1$ (equation 1)

2*3+b = -3 (equation 2)

Solve then and you get b (second equation) and c.

Originally Posted by Joel

f) the line 3x + y – 5 = 0 is a normal at the point T(3,-1)

Thanks heaps
$f(3) = 3^2 + b*3 +c = 1$ (equation 1)

because the parabola goes through T

and we further know that y = -3x+5

In this case our parabola has not the slope of y, but it is orthogonal to the slope of y.

Hence you try to find the slope f'(3), which is orthogonal to the slope m=-3 (slope of y).

So you need to solve f'(3) * m = -1 (equation 2)

f'(3) = 2*3+b

Substitute f'(3) in equation 2 and substitute m =-3

And you get

(2*3+b)*(-3) = -1 (equation 3)

You are able to find b und c by solving equation 3 and equation 1.

Yours
Rapha

4. My working out is as follows, but the answer for c is supposed to be 17.

e) the parabola is tangent to 3x + y – 5 = 0 at the point T(3,-1)

y = x² + bx + c
y’=2x + b, this is the gradient of the tangent

rearrange the equation, 3x + y – 5 = 0
y=-3x+5, the gradient is given by y=mx+b, so now gradient is known to be -3

At T(3,1), y’=2x + b
-3=2(3) + b
-3=6+b
-9=b
Back into original equation
y = x² + bx + c
1=3²+(-9)(3)+c
1=9-27+c
19=c.

Where have I gone wrong?

5. Hi.

Originally Posted by Joel
My working out is as follows, but the answer for c is supposed to be 17.

e) the parabola is tangent to 3x + y – 5 = 0 at the point T(3,-1)

y = x² + bx + c
y’=2x + b, this is the gradient of the tangent

rearrange the equation, 3x + y – 5 = 0
y=-3x+5, the gradient is given by y=mx+b, so now gradient is known to be -3

At T(3,1), y’=2x + b
-3=2(3) + b
-3=6+b
-9=b
Back into original equation
y = x² + bx + c
1=3²+(-9)(3)+c
1=9-27+c
19=c.

Where have I gone wrong?
You did everything right, except for one small mistake

You said

At T(3,1), y’=2x + b
-3=2(3) + b
-3=6+b
-9=b
No, it is not T(3,1); it should be T(3,-1). Did you notice the "minus"?

So

Back into original equation
y = x² + bx + c
-1=3²+(-9)(3)+c
-1=9-27+c
17=c.

But good job anyway!

Yours
Rapha