We have a quadratic function written in standard form, and we want to rewrite it in vertex form.
Standard Form
y= ax^2+ bx+c
Given Equation
y=- 3x^2-6x-8 ⇔ y= - 3x^2+( - 6)x+(- 8)
In the given equation, a= - 3, b= - 6, and c=- 8.
Let's now recall the vertex form of a quadratic function.
Vertex Form: y=a(x-h)^2+k
In this equation, a is the leading coefficient of the quadratic function, and the point (h,k) is the vertex of the parabola. Also, note that a vertical line with the equation x=h is the axis of symmetry. By substituting our given values for a and b into the expression - b2a, we can find h.
So far, we know that the equation of the axis of symmetry is x=- 1. Thus, the vertex lies at (- 1,k). To find the y-coordinate k, we will substitute - 1 for x in the given function.
Therefore, the (h,k) coordinate pair of the vertex is (- 1,- 5). Moreover, since we already know that a= - 3, we can rewrite the given function in vertex form.
y= - 3 (x-(- 1))^2+(- 5) ⇔ y=- 3(x+1)^2-5
