make a sketch of the graph using
x = -1,0,1,2,3
How many real number solutions does the equation have? y=3x^2-5x-5
A)one solution
B)two solutions
C)no solutions
D)infinitely many solutions
I can't figure this question out. Any help would be amazing.
11 answers
I still don't get it... Should I graph them as points? Because when I graph just x=-1 and ect... I get straight lines going up and down at the following points, (-1,0), (0,0),(1,0),(2,0),(3,0).
Is it no solutions?
No, the answer is two solutions, Correct?
your first response tells me you are having difficulties evaluating a function for a given x
x = -1 , y = 3(-1)^2 - 5(-1) - 5 = 3+5-5 = 3
so the point is (-1,3)
x = 0 , y = -5 , so the point is (0,-5)
x = 1 , y = 3(1)^2 -5(1) - 5 = -7
so the point is (1,-7)
x = 2 , y = -3 , point is (2,-3)
x = 3, y = 7 , point (3,7)
without even looking at the graph and knowing that it is a parabola opening upwards, I can see that the graph from (-1,3) to (0,-5) must have crossed the x-axis, and from (2,-3) to (3,7) must have crossed it a second time,
so, yes, there are two solutions.
Looks like you are just guessing.
x = -1 , y = 3(-1)^2 - 5(-1) - 5 = 3+5-5 = 3
so the point is (-1,3)
x = 0 , y = -5 , so the point is (0,-5)
x = 1 , y = 3(1)^2 -5(1) - 5 = -7
so the point is (1,-7)
x = 2 , y = -3 , point is (2,-3)
x = 3, y = 7 , point (3,7)
without even looking at the graph and knowing that it is a parabola opening upwards, I can see that the graph from (-1,3) to (0,-5) must have crossed the x-axis, and from (2,-3) to (3,7) must have crossed it a second time,
so, yes, there are two solutions.
Looks like you are just guessing.
2 solutions im 5 years late lol
durd
lmfao ms sue
7 years lateðŸ˜
8 years late .....
Well, at least now you know the answer if you ever come across this question again in the future!
1 answer