Question

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Answer 1

r u working with this equation for the parabola
y= a(x-h)^2 +k
?

Answer 2

ok see the equation it have 3 unkwon values
a,h,k
and in ur graph u have 3 know points, i meant 3 intercepts

Answer 3

i think we must replace that values in the equation
y= a(x-h)^2 +k
and solve to find a,h and k

Answer 4

it becomes too large 3 equations with 3 variables

Answer 5

a,h,k

Answer 6

i m trying to find another way the first way seems to large for me

Answer 7

mmm maybe u could help to solve the system of equations
for (-3,0) we got
0= a(-3-h)^2 +k
for (-1,0) we got
0= a(-1-h)^2 +k
for (0, -3) we got
-3= (0-h)^2 +k
i will come in 3 minutes

Answer 8

Okay, you have x-intercepts of -3 and -1. That means at x=-3 and x=-1, f(x) = 0. The way you achieve that is by a product:

\[y =k(x+3)(x+1)\]If you evaluate that at x=-3 or x=-1, you'll see that you get 0. The value of \(k\) is determined by your y-intercept, and is essentially a scaling factor to make the curve have the right vertical dimensions. Plug in x=0, y = -3, and solve for \(k\).

Answer 9

ohhhhhhhh a lot better

Answer 10

solution

Answer 11

if \( k = -1, y = k(x+1)(x+3) = -(x+1)(x+3) = -x^2-4x-3\)

Answer 12

In this case, the scaling factor doesn't do any scaling, only flipping the graph upside down.

Answer 13

And here's a graph