The function f(x)=x2+5x−6
is reflected over a given line and results in the equation f′(x)=−x2−5x+6
. Identify the equation for the line of reflection.
To find the line of reflection, we need to find the equation of the line that the original function, f(x) = x^2 + 5x - 6, reflects over to result in f'(x) = -x^2 - 5x + 6.
The line of reflection will have the form x = a, where a is a constant.
To determine the value of a, we can compare corresponding points on the graph of f(x) and f'(x).
The vertex of f(x) is given by (h, k) = (-b/2a, f(-b/2a)). In this case, a = 1, b = 5, and c = -6.
Therefore, h = -5/(2*1) = -5/2 and k = f(-5/2) = (-5/2)^2 + 5(-5/2) - 6 = 11/4.
The vertex of f'(x) is given by (h', k') = (-b/2a, f'(-b/2a)). In this case, a = -1, b = -5, and c = 6.
Therefore, h' = -(-5)/(2*-1) = 5/2 and k' = f'(5/2) = -(5/2)^2 - 5(5/2) + 6 = 11/4.
Since the y-coordinates of the vertices of f(x) and f'(x) are the same (k = k'), the line of reflection must be the line x = h = -5/2.
Therefore, the equation for the line of reflection is x = -5/2.
what does y equal?
We can find the equation for the line of reflection by using the formula y = mx + b, where m is the slope of the line and b is the y-intercept.
To determine the slope of the line of reflection, we can consider two points on the line: one point on the original function f(x) and the corresponding reflected point on the reflected function f'(x).
Let's choose a point on f(x) that is not on the line of reflection. For simplicity, we can choose x = 0. Therefore, the corresponding y-coordinate can be found by substituting x = 0 into the equation f(x) = x^2 + 5x - 6:
f(0) = (0)^2 + 5(0) - 6 = -6.
Now, let's find the corresponding point on f'(x). Since the line of reflection is x = -5/2, we can evaluate f'(-5/2):
f'(-5/2) = (-5/2)^2 + 5(-5/2) + 6 = 11/4.
Therefore, the two points we have are (0, -6) and (-5/2, 11/4).
To find the slope, we can use the formula:
m = (y2 - y1) / (x2 - x1),
where (x1, y1) = (0, -6) and (x2, y2) = (-5/2, 11/4):
m = (11/4 - (-6)) / (-5/2 - 0)
= (11/4 + 24/4) / (-5/2)
= 35/4 / (-5/2)
= (35/4) * (-2/5)
= -14/4
= -7/2.
Now that we have the slope of the line, we need to find the y-intercept. We can use the point-slope form of the equation:
y - y1 = m(x - x1),
where (x1, y1) = (0, -6) and m = -7/2:
y - (-6) = (-7/2)(x - 0)
=> y + 6 = (-7/2)x
=> y = (-7/2)x - 6.
Therefore, the equation for the line of reflection is y = (-7/2)x - 6.
To identify the equation for the line of reflection, we can compare the two functions, f(x) and f'(x), and analyze how the reflection affects the coefficients in each term.
Comparing the two functions:
f(x) = x^2 + 5x - 6
f'(x) = -x^2 - 5x + 6
We can observe that the quadratic term (x^2) is negated (from positive to negative) in f'(x), indicating that the line of reflection is the x-axis. When a function is reflected over the x-axis, the sign of the quadratic term changes.
Hence, the equation for the line of reflection is x-axis or y = 0.