Can you please graph the function,label the vertex and axis of symmetry of y=-8^2-12x+1

Question

lesson 10.2 graph y=ax^2+bx+c 8th grade Graphing Quadratic Functions

please show the work...thanks

Answer 1

Let y=-8x²-12x+1, so y can be written:

y=-8(x²+3x/2)+1,

y=-8(x²+3x/2+9/16-9/16)+1,

y=-8((x+¾)²-9/16)+1,

y=-8(x+¾)²+9/2+1,

y=-8(x+¾)²+11/2. This is the vertex form of the parabola.

From this the vertex is at (-¾,11/2). The axis of symmetry is the vertical line passing through the vertex, so it's x=-¾. Th parabola is "upside down" because (x+¾)² has a negative coefficient.

The vertex (-0.75,5.5) is labelled, and the axis of symmetry is the green vertical line passing through the vertex. Note also the y-intercept at 1 (when x=0, y=1).