Answer:
To graph a quadratic function with a set of {-1,3}, we need to find the equation of the function first. Since we are given two points, we can use them to form a system of equations and solve for the coefficients of the quadratic function.
Let's assume that the quadratic function has the standard form:
f(x) = ax^2 + bx + c
Using the given points (-1, 0) and (3, 0), we can set up the following system of equations:
a(-1)^2 + b(-1) + c = 0
a(3)^2 + b(3) + c = 0
Simplifying each equation, we get:
a - b + c = 0
9a + 3b + c = 0
Now we can solve this system of equations using any method we prefer. For example, we can use substitution to eliminate one of the variables. Solving for c in the first equation, we get:
c = b - a
Substituting this expression for c into the second equation, we get:
9a + 3b + (b - a) = 0
Simplifying this equation, we get:
8a + 4b = 0
Dividing both sides by 4, we get:
2a + b = 0
Solving for b in terms of a, we get:
b = -2a
Substituting this expression for b into c = b - a, we get:
c = -3a
Therefore, the quadratic function can be written as:
f(x) = ax^2 - 2ax - 3a
To find the vertex of the parabola, we can use the formula:
x = -b/2a
Substituting a = 1 and b = -2a, we get:
x = -(-2a)/(2a) = 1
To find the y-coordinate of the vertex, we can substitute x = 1 into the function f(x):
f(1) = a(1)^2 - 2a(1) - 3a = -a
Therefore, the vertex of the parabola is at the point (1, -a).
To find the x-intercepts, we can set f(x) = 0 and solve for x:
ax^2 - 2ax - 3a = 0
Dividing both sides by a, we get:
x^2 - 2x - 3 = 0
Factoring this quadratic equation, we get:
(x - 3)(x + 1) = 0
Therefore, the x-intercepts of the parabola are at x = 3 and x = -1.
To find the y-intercept, we can substitute x = 0 into the function f(x):
f(0) = a(0)^2 - 2a(0) - 3a = -3a
Therefore, the y-intercept of the parabola is at the point (0, -3a).
Finally, to find the reflection of the y-intercept in the axis of symmetry (which is x = 1), we can use the formula:
x' = 2p - x
where p is the x-coordinate of the vertex. Substituting p = 1 and x = 0, we get:
x' = 2(1) - 0 = 2
Therefore, the reflection of the y-intercept in the axis of symmetry is at the point (2, -3a).
To summarize, the quadratic function that passes through the points (-1, 0) and (3, 0) can be written as f(x) = ax^2 - 2ax - 3a, where a is any non-zero constant. The vertex of the parabola is at the point (1, -a), the x-intercepts are at x = -1 and x = 3, the y-intercept is at the point (0, -3a), and the reflection of the y-intercept in the axis of symmetry is at the point (2, -3a).