Equation of the parabola: y = ax^2 + bx + c. Find a, b, and c.
x of axis of symmetry:
x
=
−
b
2
a
=
3
-> b = -6a
Writing that the graph passing at point (1, 0) and point (4, -3):
(1) 0 = a + b + c -> c = - a - b = - a + 6a = 5a
(2) -3 = 16a + 4b + c --> -3 = 16a - 24a + 5a = -3a --> a = 1
b = -6a = -6; and c = 5a = 5
y
=
x
2
−
6
x
+
5
Check with x = 1: -> y = 1 - 6 + 5 = 0. OK
Neither. When graphing both lines they intersect but are not perpendicular.
Answer
I guess it my be 8. can you try it ?
Step-by-step explanation:
Answer:
Read this,it should help!
The standard form of a quadratic function is y = ax 2 + bx + c. where a, b and c are real numbers, and a ≠ 0. Using Vertex Form to Derive Standard Form. Write the vertex form of a quadratic function. y = a(x - h) 2 + k. Square the binomial. y = a(x 2 - 2xh + h 2) + k. y = ax 2 - 2ahx + ah 2 + k
