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Thread: sketching of Parabola graph

  1. #1
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    Exclamation sketching of Parabola graph

    how do I draw this? What method can I use?

    Q: Plot the graph of y=3(x+2)squared -3 for the domain -5<x<2 and determine the following:
    1.1 Turning point
    1.2 Axis of symmetry
    1.3 x-intercepts
    1.4 y-intercepts
    1.5 domain
    1.6 Range
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  2. #2
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    Quote Originally Posted by Dreamer View Post
    how do I draw this? What method can I use?

    Q: Plot the graph of y=3(x+2)squared -3 for the domain -5<x<2 and determine the following:
    1.1 Turning point
    1.2 Axis of symmetry
    1.3 x-intercepts
    1.4 y-intercepts
    1.5 domain
    1.6 Range
    $\displaystyle P: y = 3(x+2)^2-3$ . With such a form of the equation of a parabola you can get all necessary values.

    1.1 Turning point: V(-2, -3)
    1.2 Axis of symmetry: x = -2. The parabola opens upward (3>0) and is stretched(?) by the factor 3.
    1.3 x-intercepts: $\displaystyle y = 0~\implies~(x+2)^2=1~\implies~x=-3~\vee~x=-1$
    1.4 y-intercepts: $\displaystyle x = 0~\implies~y=9~\implies~Y(0,9)$
    1.5 domain: given: $\displaystyle -5 < x < 2$
    1.6 Range:$\displaystyle min_p = -3$ see coordinates of vertex.
    $\displaystyle p(-5) = 24$, $\displaystyle p(2) = 45$. Therefore $\displaystyle max_p = p(2)
    $
    Therefore the range is r = (-3;45)
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