1.) Determine the equation of the axis of symmetry.
y = -5x2 - x + 9
Answer 1.) x=3/2
2.) Determine the equation of the axis of symmetry.
y = -6x2 + 3x - 4
Answer 2.) x=-2/10
I assume you have not done calculus before.
To find the axis of symmetry of a polynomial of second degree (quadratic), we only have to complete the squares, which will then give the coordinates of the vertex in the form V(h,k) where
y=a(x-h)²+k
and a is another constant.
Starting with
y = -5x2 - x + 9
we write
y=-5(x²+2(x/10))+9
=-5(x+1/10)²+5/100 +9
=-5(x+1/10)² +9.05
Therefore h=-1/10, k=9.05, and
V(-1/10,9.05)
or the axis of symmetry is x=-1/10
I will leave #2 for you as practice.
Thanks! so number two is wrong?
#2 is not correct.
You can follow the steps above and try again.
To determine the equation of the axis of symmetry, we need to find the x-coordinate of the vertex of the parabola.
The equation of a parabola in vertex form is y = a(x-h)^2 + k, where (h, k) represents the coordinates of the vertex.
To find the x-coordinate of the vertex, we use the formula x = -b/2a, where a, b, and c are the coefficients of the quadratic equation in standard form (ax^2 + bx + c = 0).
1.) For the equation y = -5x^2 - x + 9:
a = -5, b = -1, and c = 9
Using the formula x = -b/2a, we substitute the values:
x = -(-1) / (2 * (-5))
x = 1/(-10)
x = -1/10
So, the x-coordinate of the vertex is -1/10.
Hence, the equation of the axis of symmetry is x = -1/10.
Similarly, for the equation y = -6x^2 + 3x - 4:
a = -6, b = 3, and c = -4
Using the formula x = -b/2a, we substitute the values:
x = -3 / (2 * (-6))
x = -3 / (-12)
x = -(-1/4)
x = 1/4
So, the x-coordinate of the vertex is 1/4.
Hence, the equation of the axis of symmetry is x = 1/4.