Axis of symmetry(if exists) goes through the vertex and is parallel to the y axis. The axis of symmetry of considered quadratic function is Its vertex is at point (-2,4) . Its equation is given as: [tex]y = -x^2 - 4x[/tex]
How to find the equation of a quadratic equation passing through three points?
The general form of a quadratic function is: [tex]y = ax^2 + bx + c[/tex]
This contains three variables. Using 3 points will give us values of these three constants (a, b and c), thus, providing us the needed equation. It is shown below how we can do that.
For the given case, let the considered quadratic function be [tex]y = ax^2 + bx + c[/tex]
Then, as its graph passes through (−3, 3), (−1, 3), and (−2, 4), thus, these points will satisfy the aforesaid equation.
Since points are in the form (x,y), thus,
[tex]y = ax^2 + bx + c\\\\3 = a(-3)^2 + b(-3) + c = 9a -3b + c\\\\3 = a(-1)^2 + b(-1) + c = a - b + c\\\\4 = a(-2)^2 + b(-2) + c = 4a -2b + c[/tex]
And therefore we get a system of three equations as:
[tex]9a - 3b + c = 3\\a - b + c = 3\\4a - 2b + c = 4[/tex]
Subtracting equation second from first, we get:
[tex]9a - a -3b + b = 0\\8a = 2b\\4a = b[/tex]
Subtracting equation second from third equation, we get:
[tex]4a - a -2b + b = 4-3\\3a = b + 1\\ 3a-1 = b[/tex]
Thus, from the two later equations we got,
[tex]4a = b = 3a-1\\4a = 3a-1\\a = -1[/tex]
Thus, we get value of b as : b = 4a = -4
Using these two values, and putting them in the first equation of the three linear equations, we get:
[tex]9a - 3b + c = 3\\-9 + 12 + c = 3\\c = 0[/tex]
Thus, the quadratic equation which passes through given points is:
[tex]y= ax^2 + bx + c = -x^2 -4x[/tex]
What is the vertex form of a quadratic function?
The form [tex]y = a(x-h)^2 + k[/tex] quadratic function. Its vertex is at point (h, k)
Its axis of symmetry (if function is symmetric) is given by x = h
Converting the obtained function to vertex form, we get:
[tex]y = -x^2 - 4x\\y = -(x^2 + 4x + 4-4) = -((x+2)^2 - 4)\\y = -(x+2)^2 + 4[/tex]
Comparing it with the vertex form, we get h = -2, and k = 4, Thus, its vertex lies on (h,k) = (-2, 4)
The axes of symmetry passes through vertex point (h,k) and is parallel to y-axis, that means its equation is of the form x = h)
Thus, axis of symmetry for it is: x = -2 (since a line parallel to y axis is of the form x = a, where a is its intersection with x axis)
The graph of the considered quadratic function is given below.
Thus, the axis of symmetry of considered quadratic function is Its vertex is at point (-2,4) . Its equation is given as: [tex]y = -x^2 - 4x[/tex]
Learn more about vertex form of a quadratic equation here:
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