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anonymous

  • 5 years ago

find rules for quadratic functions with graphs meeting these conditions. 1.x-intercepts at (4,0)and(-1,0) and opening upward 2.x-intercepts at (2,0)and(6,0) and maximum point=(4,12) 3.only one x-intercept and that=(-3,0) and the y-intercept=(0,18)

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  1. anonymous
    • 5 years ago
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    in general, a quadratic equation is in the form ax^2 + bx + c = 0. So from the conditions given, you have to determine the value of a, b and c.

  2. anonymous
    • 5 years ago
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    1. the function has x-intercepts at (4,0)and(-1,0), so we can simply write the function as f(x)=a(x-4)(x+1). also, it is mentioned that the function has an opening upward, so a is required to be a positive value

  3. anonymous
    • 5 years ago
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    2.in the same way of number 1, we can write this function as f(x)=a(x-2)(x-6). notice that (4,12) is on the graph of the function, and therefore f(x) satisfies that a(4-2)(4-6)=12, so a=-3. consequently, the function is f(x)=-3(x-2)(x-6)=-3x^2+24x-36

  4. anonymous
    • 5 years ago
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    1.opening upward, means a>0. It crosses two points (4,0) and (-1,0), so we can write y = a(x-4)(x+1) 2. y = a(x-2)(x-6) because it passes the points (2,0) and (6,0). Since it has maximum, then a<0 and since the maximum happens at (4,12), we can determine a by solving a(4-2)(4-6) = 12. 3. The quadratic graph only touch the x-intercept at (-3,0), so it has the form y = a(x+3)^2. Then it also passes through the point (0,18), we can determine a by solving 18 = a(0+3)^2

  5. anonymous
    • 5 years ago
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    another solution for 3: why not set the function as f(x)=ax^2+bx+18 because of the y-intercept. now we have to imagine the that (-3,0) should be the lowest point of the graph of the function. so -b/2a=-3, (4a*18-b^2)/4a=0. a little bit more difficult, but ok as well. a=2, b=12. f(x)=2x^2+12x+18

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