Hi, the question is: Write the equation of the parabola with vertex(3,-1) and x-intercepts 2 and 4.
I did this:
y=?(x-3)^2-1
How to find the a-value?
Many thanks, and I appreciate any answer!
y=a(x-3)^2-1
x-intercept 2 ----> the point (2,0) lies on it
0 = a(1) - 1
a = 1
y=(x-3)^2-1
To find the "a" value in the equation of a parabola, you can use the vertex form of the equation, which is given by:
y = a(x - h)^2 + k
In this equation, the vertex is represented by the point (h, k). So in your case, the vertex is (3, -1). Substituting these values into the equation, we have:
y = a(x - 3)^2 - 1
As you correctly identified.
To find the "a" value, we can use the x-intercepts of the parabola. The x-intercepts occur when y = 0. So we can substitute y with 0 in the equation.
0 = a(x - 3)^2 - 1
Rearranging the equation, we get:
1 = a(x - 3)^2
Now we can substitute one of the x-intercepts into the equation. Let's use x = 2:
1 = a(2 - 3)^2
1 = a(-1)^2
1 = a(1)
1 = a
Therefore, the value of "a" is 1.
Now we can update the equation:
y = (x - 3)^2 - 1
So the final equation of the parabola with a vertex (3, -1) and x-intercepts 2 and 4 is:
y = (x - 3)^2 - 1
Hope this helps! Let me know if you have any other questions.