it's easy you first add 19 to the and then complete the square with x^2 that would be (x-4)^2 and then you find the mid point which is (4,0),well not mid point but vertex and graph it.
Elementary, the parabola does have its Vertex at ,as you say (2, 0) but you see it is up side down Always check by using x =0 (here f(x)=-8)and also at symm opposite point x = 4 f(x )=-8 and at x = 1 & at x =3 it has same value -2 So cheerup nothing wrong only lack of knowledge dear ok?
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it's easy you first add 19 to the and then complete the square with x^2 that would be (x-4)^2 and then you find the mid point which is (4,0),well not mid point but vertex and graph it.
hope it helps
Elementary, the parabola does have its Vertex at ,as you say (2, 0) but you see it is up side down Always check by using x =0 (here f(x)=-8)and also at symm opposite point x = 4 f(x )=-8 and at x = 1 & at x =3 it has same value -2 So cheerup nothing wrong only lack of knowledge dear ok?
Are you sure that is Algebra 2 cuz i am in Algebra 1 and we are doing parabolas.
WELL Y= -19 NEGATIVE 19... SO SINE HAVE THAT ALL YOU HAVE TO DO IS WORK THE PROBLEM BACKWARDS THEN U WILL GET X. OK
it is unsolvable!!!
19 doesnt factor into 8...u hafta use the quadratic formual.....b2-(squareroot of 4AC) divided by 2A
i really dont think u can do it but i really dont have a clue!