Question

BRAINLIESTTT ASAP! PLEASE HELP ME :)

Answer 1

Answer:

Here's what I get  

Step-by-step explanation:

1. Graph the parabola

(a) Create a table containing a few values of x and y

I chose the following points:

$\begin{array}{rr}\mathbf{x} & \mathbf{f(x)}\\-3 & 0\\-2 & -3 \\-1 & -4\\0 & -3\\1 & 0 \\2 & 3\\\end{array}$

(b) Plot your points

Draw dots at the coordinates of each point ( Fig. 1).

(c) Draw the graph

Draw a smooth line through the points.

Extend your line smoothly to the edge of the plot area (Fig. 2).

2. Label the intercepts and vertex

The x-intercepts are at (-3, 0) and (1, 0).

The y-intercept is at (0, -3).

The vertex is at (-1, -4).

See Fig.3 for the labels.

3. Axis of symmetry

The axis of symmetry is a vertical line that passes through the vertex and divides the graph into mirror- image halves.

The vertex is at (-1, -4), so the vertex is a straight line passing through (-1, -4) and (-1, 0). It is the dashed black line in Fig. 4

Answer 2

x-intercepts: (-3,0), (1,0)

work:

0 = x^2 + 2x - 3

quadratic equation: x = -2 +√(2^(2) - 4 × 1(-3)) / (2 × 1) = 1

quadratic equation: x = -2 -√(2^(2) - 4 × 1(-3)) / (2 × 1) = -3

(x,y) --) (-3,0), (1,0)

y-intercept: (0,-3)

work:

y = (0)^2 + 2(0) - 3

y = -3

(x,y) --) (0,-3)

vertex: (-1,-4)

work:

xv = -( b / (2a) )

a = 1, b = 2, c = -3

xv = -( 2 / (2×1) )

xv = -1

yv = (-1)^2 + 2(-1) - 3

yv = -4

(x,y) --) (-1,-4)

axis of symmetry: -1

work:

a = 1 in the x^(2) + 2ax + a^(2)

x^(2) + 2x + 1^(2) = (x + 1)^(2)

(x + 1)^(2) - 3 - 1^(2)

y = (x + 1)^(2) - 4

y + 4 = (x - 1)^(2)

put in standard form --) 4 × 1/4( y -( -4 ) ) = ( x -( -1) )^(2)

(h,k) = (-1,-4); p = 1/4

in parabola form expression: 4p( y - k ) = ( x - h )^(2) and is symmetric around the y-axis at -1.