how to find the points thought which the function goes like if the vertex is 2,5 and it goes through (6,-1) how would i find the other point?
To find the other point through which the function passes, we need to determine the equation of the function.
Let's assume the equation of the function is a quadratic of the form y = ax^2 + bx + c, where (h, k) is the vertex. In this case, (2,5) is the vertex of the quadratic function.
Using the vertex form of the quadratic function, we can write the equation as y = a(x - h)^2 + k. Substituting the vertex coordinates (2,5) into the equation, we get y = a(x - 2)^2 + 5.
Now, we use the given point (6,-1) to find the value of 'a'. We substitute the x and y values of the point into the equation and solve for 'a'.
-1 = a(6 - 2)^2 + 5
-1 = a(4)^2 + 5
-1 = 16a + 5
16a = -1 - 5
16a = -6
a = -6/16
a = -3/8
Now that we have the value of 'a', we can substitute it back into the equation y = a(x - 2)^2 + 5. Hence, the equation of the quadratic function is y = (-3/8)(x - 2)^2 + 5.
To find the other point through which the function passes, we can substitute any arbitrary x-value into the equation and solve for y. Let's choose a different x-value, say x = 0. Plugging this into the equation, we get:
y = (-3/8)(0 - 2)^2 + 5
y = (-3/8)(-2)^2 + 5
y = (-3/8)(4) + 5
y = -3/2 + 5
y = -3/2 + 10/2
y = 7/2
Therefore, the other point through which the function passes is (0, 7/2).