I would like to have some help on this problem please. I'm so post to find the vertex and axis of the parabola.Does anyone have an idea?
Y=x²-16x+63.
A)vertex(8,-1)axis=x=8
B)vertex(8,1)axis=x=8
C)vertex(-8,-1)axis=x=-8
D)vertex(-8,1)axis=x=-8
x^2-16x= y-63
x^2 - 16x + 64 = y+64-63
(x-8)^2 = y+1
symmetric about the vertical axis x = 8
when x = 8, y = -1
To find the vertex and axis of a parabola in the form of y = ax^2 + bx + c, we can use the formula:
x = -b/2a
In the given equation y = x² - 16x + 63, we can identify that a = 1, b = -16, and c = 63.
1. Start by substituting the values of a and b into the formula: x = -(-16) / (2 * 1)
Simplifying, we have: x = 16 / 2
2. Now, compute the value of x: x = 8
3. After finding the value of x, substitute it back into the equation to find the y-coordinate:
y = (8)² - 16(8) + 63
y = 64 - 128 + 63
y = -1
Therefore, the vertex of the parabola is (8, -1).
4. To find the axis of the parabola, we use the x-coordinate of the vertex, which is 8.
Thus, the correct answer is A) vertex(8,-1) axis=x=8.