the quadratic function (1 Viewer)

jnney · Aug 18, 2011

find the gradients of the tangents to y=2+5x-3x^2 which pass through the point (1,7)

Alkanes · Aug 18, 2011

dy/dx = 5 - 6x
= 5 - 6(1)
= -1??

powlmao · Aug 18, 2011

$y=2+5x=3x^2$
$\frac{dx}{dy}=5-6x$
When x=1
$5-6, mt=-1$
$y-y1=m(x-x1)$
Sub in
$y-7=-1(x-1)$
y+x-8=0

Whoops, found the equation of the tangent.

jnney · Aug 18, 2011

The answer is -7 and 5.

And by the way, you can't just differentiate it and sub x = 1 in , because that coordinate doesn't like on the curve.

powlmao · Aug 18, 2011

jnney said:
The answer is -7 and 5.

And by the way, you can't just differentiate it and sub x = 1 in , because that coordinate doesn't like on the curve.

Argh, i messed up and clicked the wrong buttom

Alkanes · Aug 18, 2011

I knew something was suss haha

powlmao · Aug 18, 2011

Besides the number typo I have the method right?

elbatiolpxeho · Aug 18, 2011

The general equation of any line through (1,7) is:

$\\y-7=m(x-1)\\y=mx+(7-m)$

Sub into other function:

$\\mx+(7-m)=2+5x-3x^2\\3x^2+(m-5)x+(5-m)=0$

If the line is a tangent then the discriminant is equal to zero.

$\\\Delta=0$

Solve for m then go fuck bitches mmmkay.

SpiralFlex · Aug 18, 2011

powlmao said:
$y=2+5x=3x^2$
$\frac{dx}{dy}=5-6x$
When x=1
$5-6, mt=-1$
$y-y1=m(x-x1)$
Sub in
$y-7=-1(x-1)$
y+x-8=0

Whoops, found the equation of the tangent.

You are differentiating with respect to $x$ . So it should be $\frac{dy}{dx}.$ As the title suggest you would involve the use of discriminants.

the quadratic function (1 Viewer)

jnney

lemon

Alkanes

Active Member

powlmao

Banned

jnney

lemon

powlmao

Banned

Alkanes

Active Member

powlmao

Banned

elbatiolpxeho

Member

SpiralFlex

Well-Known Member

Users Who Are Viewing This Thread (Users: 0, Guests: 1)