1. ## help with function

heyy here is a question i have, i was wondering if anyone could help me:

The function f is defined by f:x --> -0.5x^2 + 2x + 2.5

a. Let N be the normal to the curve at the point wehre the graph intercpets the y-axis. Show that the equation of N may be written as y = -0.5x + 2.5

Let g:x --> -0.5x + 2.5

b. Find the solutions of f(x) = g(x)
c. Hence find the coordinates of the other point of intersection of the normal and the curve.

Thanks!

2. [quote=samantha_malone;60972]heyy here is a question i have, i was wondering if anyone could help me:

The function f is defined by f:x --> -0.5x^2 + 2x + 2.5

a. Let N be the normal to the curve at the point wehre the graph intercpets the y-axis. Show that the equation of N may be written as y = -0.5x + 2.5

Let g:x --> -0.5x + 2.5

[/tex]

The curve intercepts the y-axis when $x=0$, and so $y=2.5$. The slope of the curve is:

$
\frac{d}{dx}(-0.5x^2 + 2x + 2.5)=-x+2
$

so at the intercept the slope is $2$, and hence the slope of the normal is $-0.5$. So the normal has slope $-0.5$ and passess through $(0,2.5)$, and so its equation is:

$
y=0.5x+2.5
$

as required.

RonL

3. a) $f(0)=2.5$, so the point where the graph intercepts the y-axis has the coordinates $(0,2.5)$.
The normal to the graph in this point is perpendicular to the tangent in the same point.
The slope of the tangent is $f'(0)$.
We have $f'(x)=x+2\Rightarrow f'(0)=2$.
The slope of the normal is $m=-\frac{1}{2}=-0.5$.
The equation of the normal is $y-y_0=m(x-x_0)$, where $x_0=0,y_0=2.5,m=-0.5$.
Then the equation is $y-2.5=-0.5x\Leftrightarrow y=-0.5x+2.5$

b) $f(x)=g(x)\Leftrightarrow 0.5x^2+2x+2.5=-0.5x+2.5\Leftrightarrow$
$\Leftrightarrow 0.5x^2+2.5x=0\Leftrightarrow x(0.5x+2.5)=0\Rightarrow$
$\Rightarrow x_1=0,x_2=-5$

c) $f(-5)=5$, so the second point where the normal intercepts the graph has the coordinates $(-5,5)$.

4. thanks!