How do you find the axis of symmetry, vertex and x intercepts for #y=-x^2-2x#?

Question

Eva Abramson · Accepted Answer

Axis of sym at #x= -1#, vertex at (-1,1)

#x-#intercepts at -2 and 0

#y#-intercept at (0,0)

Let's find the axis of symmetry first, it is an easy way to find the vertex. #color(red)("axis of symmetry:" x= (-b)/(2a)) = (-(-2))/(2(-1)) = -1# #x=-1# The vertex lies on the axis of symmetry: If #x=-1, " find " y = -(-1)^2 -2(-1) = -1+2 = 1# #color(blue)("The vertex is at " (-1,1))# Intercepts: For y-int: #x=0 rarr y=0# For x-intercepts : # y = 0# #0 = -x^2 -2x" "larr # factorise #0 = -x(x+2) rArr x=0, or x=-2# ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ Using the completing the square method to find the vertex: #y = -x^2 -2x# #y = -1(x^2 +2x)# #y=-1[color(magenta)(x^2 +2x +1)-1]# #y =-1[color(magenta)((x+1)^2) -1]# #y = -(xcolor(blue)(+1))^2 color(blue)(+1)# #color(blue)("The vertex is at " (-1,1))# Confirm these points on the graph below. graph{-x^2-2x [-3.5, 1.5, -1.4, 1.1]}

Eva Abramson · Answer

undefined To find the axis of symmetry, vertex, and x-intercepts for the equation $y = -x^2 - 2x$:

1. Axis of symmetry: Use the formula $x = -\frac{b}{2a}$, where $a$ and $b$ are the coefficients of the quadratic equation in the form $ax^2 + bx + c$. In this case, $a = -1$ and $b = -2$. Plug these values into the formula to find the axis of symmetry.

2. Vertex: Once you have the x-coordinate of the axis of symmetry, substitute it into the original equation to find the corresponding y-coordinate.

3. X-intercepts: To find the x-intercepts, set $y$ equal to zero and solve the quadratic equation $y = -x^2 - 2x$ for $x$.