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Thread: help with function

  1. #1
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    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!
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  2. #2
    Grand Panjandrum
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    [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 $\displaystyle x=0$, and so $\displaystyle y=2.5$. The slope of the curve is:

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

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

    $\displaystyle
    y=0.5x+2.5
    $

    as required.

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

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

    c) $\displaystyle f(-5)=5$, so the second point where the normal intercepts the graph has the coordinates $\displaystyle (-5,5)$.
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  4. #4
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    thanks!
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