Write an equation of the quadratic function f whose graph has x-intercepts 3 and 7 and f(5) = 8. I tried foiling the two zeros but that doesn't give me 8 when for f(5). Like: (x -3)(x - 7) x² - 10x + 21 (5)² - 10(5) + 21 = -4 ?
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Hello Originally Posted by Power Write an equation of the quadratic function f whose graph has x-intercepts 3 and 7 and f(5) = 8. I tried foiling the two zeros but that doesn't give me 8 when for f(5). Like: (x -3)(x - 7) x² - 10x + 21 (5)² - 10(5) + 21 = -4 ? This was a good trial You're just forgetting something : the leading coefficient. a(x-3)(x-7) also has x-intercepts at 3 & 7 , where a is a constant (different to 0, if possible ^^) Then use f(5)=8 to determine a.
0 = a(x - 3)(x - 7) 8 = a(5 - 3)(5 - 7) -2 = a ?? I dont think thats right. Please help further.
Originally Posted by Power y = a(x - 3)(x - 7) 8 = a(5 - 3)(5 - 7) -2 = a ?? I dont think thats right. Please help further. Why do you think this isn't right ? It's exactly the solution Though you can expand -2(x-3)(x-7) in order to get a quadratic form... You can choose whatever display you want^^
Originally Posted by Power 0 = a(x - 3)(x - 7) 8 = a(5 - 3)(5 - 7) -2 = a ?? I dont think thats right. Please help further. What's the problem?
I got it, I just made an arithmetic mistake while checking my answer.
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