I'm at such a loss.

In exercise 4 I don't even know where to being, and in exercise 9 the x=a confuses me.

Thanks in advance

Printable View

- Oct 29th 2007, 07:28 AMFnusDerivatives
I'm at such a loss.

In exercise 4 I don't even know where to being, and in exercise 9 the x=a confuses me.

Thanks in advance - Oct 29th 2007, 09:48 AMSean12345
4a) We have $\displaystyle 2x+y=6; x=1 \implies y=4$

Also we know the gradient; $\displaystyle m=-2$

Differentiate $\displaystyle y=x^2+ax+b$

$\displaystyle \implies \frac {dy}{dx} = 2x+a$

$\displaystyle \implies 2x+a=-2$

$\displaystyle x=1; 2(-1)+a=-2$

$\displaystyle \implies a=0$

$\displaystyle \implies y=x^2+b$

$\displaystyle x=1, y=-4; -4=1^2+b$

$\displaystyle \implies b=-5$

b is similar so ill let you try.