y= -3x^2 -4x + 6 What are the vertex, axis of symmetry, maximum or minimum value, and range? Can someone help me please? Thanks!

1 Answer
Sep 10, 2017

Axis of symmetry: x=-2/3

Vertex: (-2/3,22/3) larr maximum point of the parabola

Range: {yinRR:color(white)(.)y<=22/3}

Explanation:

Given:

y=-3x^2-4x+6 is a quadratic equation in standard form,

y=ax^2-bx+c,

where:

a=-3, b=-4, and c=6

Axis of Symmetry: value of the vertical line x in which the parabola will be divided into two equal halves; also the x value for the vertex.

x=(-b)/(2a)

x=(-(-4))/(2*-3)

x=-4/6

x=-2/3 larr x axis of symmetry and x value of the vertex.

Vertex: minimum or maximum point, (x,y), of a parabola. If a>0, the vertex is the minimum point and the parabola will open upward. If a<0, the vertex is the maximum point and the parabola will open downward.

To determine the y value of the vertex, substitute -2/3 for x and solve for y. Recall that a whole number, n is a fraction of n/1.

y=-3/1(-2/3)^2-4(-2/3)+6

y=-3/1(4/9)+8/3+6

The least common denominator is 9. Multiply 8/3 by color(teal)(3)/color(teal)(3), and 6/1xxcolor(magenta)(9)/color(magenta)9 to make all numbers have the same (common) denominator.

y=-12/9+8/3xxcolor(teal)(3)/color(teal)(3)+6/1xxcolor(magenta)(9)/color(magenta)9

y=-12/9+24/9+54/9

y=((-12+24+54))/9

y=66/9

y=22/3

Vertex: (-2/3,22/3) larr maximum point of the parabola

Approximate vertex: (-0.667,7.33)

Range: all real numbers for y that are part of the parabola.

Range: {yinRR:color(white)(.)y<=22/3}

graph{y=-3x^2-4x+6 [-10, 10, -5, 5]}