Answer
The equation of the quadratic function that passes through (0, 1) and has a vertex at (1,4) is
y = -3x² + 6x + 1
Step-by-step explanation
The general form of a quadratic equation is
y = ax² + bx + c
We are told that the function passes through (0, 1) and has its vertex (the highest or lowest point on the graph) at (1, 4).
We can obtain simultaneous equations by substituting these into the general form of a quadratic equation.
(0, 1)
when x = 0, y = 1
y = ax² + bx + c
1 = a(0²) + b(0) + c
1 = 0 + 0 + c
c = 1 ...... equation 1
(1, 4)
y = ax² + bx + c
4 = a(1²) + b(1) + c
4 = a + b + c
But c = 1, So
4 = a + b + 1
a + b = 4 - 1
a + b = 3 ....... equation 2
Then, the third one, at the vertex, the first derivative of the function is 0.
y = ax² + bx + c
(dy/dx) = 2ax + b = 0
At this point, x = 1
2ax + b = 0
2a(1) + b = 0
2a + b = 0 ...... equation 3
Writing these equations together
c = 1
a + b = 3
2a + b = 0
From equation 3, b= -2a
Substituting this into equation 2
a + b = 3
a - 2a = 3
-a = 3
a = -3
b = -2a = -2 (-3) = 6
So,
a = -3
b = 6
c = 1
y = -3x² + 6x + 1
Hope this Helps!!!