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anonymous
 5 years ago
find rules for quadratic functions with graphs meeting these conditions.
1.xintercepts at (4,0)and(1,0) and opening upward
2.xintercepts at (2,0)and(6,0) and maximum point=(4,12)
3.only one xintercept and that=(3,0) and the yintercept=(0,18)
anonymous
 5 years ago
find rules for quadratic functions with graphs meeting these conditions. 1.xintercepts at (4,0)and(1,0) and opening upward 2.xintercepts at (2,0)and(6,0) and maximum point=(4,12) 3.only one xintercept and that=(3,0) and the yintercept=(0,18)

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0in 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.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.01. the function has xintercepts at (4,0)and(1,0), so we can simply write the function as f(x)=a(x4)(x+1). also, it is mentioned that the function has an opening upward, so a is required to be a positive value

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.02.in the same way of number 1, we can write this function as f(x)=a(x2)(x6). notice that (4,12) is on the graph of the function, and therefore f(x) satisfies that a(42)(46)=12, so a=3. consequently, the function is f(x)=3(x2)(x6)=3x^2+24x36

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.01.opening upward, means a>0. It crosses two points (4,0) and (1,0), so we can write y = a(x4)(x+1) 2. y = a(x2)(x6) 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(42)(46) = 12. 3. The quadratic graph only touch the xintercept 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

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0another solution for 3: why not set the function as f(x)=ax^2+bx+18 because of the yintercept. 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*18b^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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