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bridgetolivares

  • 3 years ago

Describe the differences between the graph of y = –3(x + 7)2 – 10 and the standard position graph of y = x2.

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  1. Brent0423
    • 3 years ago
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    please indicate whether its a 2 or a squared "(x+7)2"

  2. bridgetolivares
    • 3 years ago
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    yes it is squared

  3. zepp
    • 3 years ago
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    The parabola could be written as \(y = a(b(x-h)^2 + k\) Where, a,b,h,k are some constants.

  4. zepp
    • 3 years ago
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    a describes the vertical stretch; b describes the horizontal stretch; (h,k) is the vertex.

  5. zepp
    • 3 years ago
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    In your parabole, can you identify a,b,h and k for me? :)

  6. zepp
    • 3 years ago
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    parabola*

  7. Brent0423
    • 3 years ago
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    y = –3(x + 7)^2 – 10 y=-3(x^2+49)-10 y=-3x^2-147-10 y=-3x^2-157 since the 3 is negative it indicate that the graph opens downward *NOTE* if the number before the x^2 is negative the graph opens downwards, if it is positive then it opens upwards the graph y=x^2 isnt this just y=1x^2 1 is positive so the parabola opens upwards INSTEAD of downwards.

  8. zepp
    • 3 years ago
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    @Brent0423 (x+7)^2 doesn't give x^2 + 49, it gives x^2 + 14x + 49.

  9. zepp
    • 3 years ago
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    As \((a+b)^2\ = a^2 + 2ab+b^2\)

  10. Brent0423
    • 3 years ago
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    oops thought i put that, sorry

  11. Brent0423
    • 3 years ago
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    you can still see the difference in the graphs just by looking at the number in front of the x^2

  12. zepp
    • 3 years ago
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    Let me find a,b,h,k for you :) y = –3(x + 7)^2 – 10 -> y = –3(x - (-7))^2 – 10 (Notice the negative!) -3 would be a, the vertical stretch 1 would be b, since there's nothing in front of x Vertex would be at (-7, -10) Now let's take a look at our basic parabola, y = x^2 a = 1 b = 1 Vertex: (0,0) Since our a constant is a negative number AND greater than 1, we can say that this function has been stretch vertically of factor 3. Then b is the same. Vertex (0,0) and (-7,-10) We can say that the parabola is moved of 7 to the left and 10 downward.

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