Question 35
Find the axis of symmetry of the following quadratic equation. Answer with x=
y=-x^2-4x+1
To find the axis of symmetry of the quadratic equation y=-x^2-4x+1, we can use the formula x = -b/(2a), where a is the coefficient of x^2 and b is the coefficient of x.
In this case, the coefficient of x^2 is -1 and the coefficient of x is -4. Plugging these values into the formula, we get:
x = -(-4)/(2*(-1))
x = 4/-2
x = -2
Therefore, the axis of symmetry of the quadratic equation y=-x^2-4x+1 is x = -2.
Question 36 Find the vertex of the following quadratic equation. Answer as an ordered pair (x,y)
y=2x^2+8
To find the vertex of the quadratic equation y=2x^2+8, we can first rewrite the equation in vertex form. The general form of a quadratic equation in vertex form is y = a(x-h)^2 + k, where (h,k) is the vertex of the parabola.
Given equation: y = 2x^2 + 8
Factor out the 2: y = 2(x^2 + 4)
Now, to make x^2 + 4 a perfect square, we complete the square:
y = 2(x^2 + 4) = 2(x^2 + 2x + 1 - 1) = 2(x+1)^2 - 2
Comparing this to the vertex form, we see that the vertex (h,k) is (-1, -2).
Therefore, the vertex of the quadratic equation y=2x^2+8 is (-1, -2).