Asked by June
What is the solution to the quadratic equations in the graphs linked? Thanks!
1) imgur.com/4Vm7uQ0
2) imgur.com/VcEybYO
3) imgur.com/gCqO8iE
1) imgur.com/4Vm7uQ0
2) imgur.com/VcEybYO
3) imgur.com/gCqO8iE
Answers
Answered by
Reiny
look for integer valued points, especially x-intercepts
In the first, I see (1,0) and (-6,0)
so the equation must be of the form
y = a(x - 1)(x + 6) = a(x^2 + 5x - 6)
another integer point seems to be (-4,-10), so sub that into my equation:
-10 = a(-4-1)(-4+6)
-10 = a(-10)
a = 1
so y = (x-1)(x+6) or y = x^2 + 5x - 6
do the others in the 2nd in the same way
For the third, it looks like the x-intercepts are 3 and -2.5 , and another integer point is (-2,-10)
proceed in the same way
In the first, I see (1,0) and (-6,0)
so the equation must be of the form
y = a(x - 1)(x + 6) = a(x^2 + 5x - 6)
another integer point seems to be (-4,-10), so sub that into my equation:
-10 = a(-4-1)(-4+6)
-10 = a(-10)
a = 1
so y = (x-1)(x+6) or y = x^2 + 5x - 6
do the others in the 2nd in the same way
For the third, it looks like the x-intercepts are 3 and -2.5 , and another integer point is (-2,-10)
proceed in the same way
Answered by
June
Can I have answers for the other two as well? If that's not a bother, of course. I've accepted that I can never understand this stuff well enough to solve it. Thanks!
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