Please!! Do this now!

Use the function f(x) = 4x2 + 8x − 5 to answer the questions.

(Already did Part A)

Part B: What are the x-intercepts of the graph of f(x)? Show your work. (2 points)

Part C: Describe the end behavior of the graph of f(x). Explain. (2 points)

Part D: What are the steps you would use to graph f(x)? Justify that you can use the answers obtained in Part B and Part C to draw the graph. (4 points)

Part B: x = -5/2, and x = 1/2,

Part C: f(x)→ + ∞, as x→ − ∞, f(x)→ + ∞, as x→ + ∞ ,

Part D: I would start with determining the vertex. Knowing it's properties ( f(x)→ + ∞, as x→ − ∞ etc. ) we would determine the type of graph, and provided it's x - intercepts we can plot two points on either side and connect them to this vertex and extend beyond infinitely

Step-by-step explanation:

Part B: It would be easier to factor the expression 4x^2 + 8x - 5, firstly;

( 4x^2 - 2x ) + ( 10x - 5 ),

2x ( 2x - 1 ) + 5 ( 2x - 1 ),

Factored Expression: ( 2x + 5 )( 2x - 1 ) ⇒ f ( x ) = ( 2x + 5 )( 2x - 1 )

Now this equation isn't in standard form, but to determine the x - intercept the y - value must still be 0;

0 = ( 2x + 5 )( 2x - 1 ),

2x + 5 = 0, and 2x - 1 = 0,

x = -5/2, and x = 1/2

Part C: Knowing that this graph extends upwards ( an upward facing parabola ), the end behavior would be such;

f(x)→ + ∞, as x→ − ∞,          ⇒        demonstrated by the above

f(x)→ + ∞, as x→ + ∞        

Part D: I would start with determining the vertex. Knowing it's properties ( f(x)→ + ∞, as x→ − ∞ etc. ) we would determine the type of graph, and provided it's x - intercepts we can plot two points on either side and connect them to this vertex and extend beyond infinitely

for the friend request.

when a figure, such as triangle (xyz) is reflected over the line y = x to create x'y'z' the x-coordinates and y-coordinates of the original image (pre-image) will be reversed for the image. (e.g., (x, y) > (y, x)) this means that the pre-image position of the x- and y-coordinates switch places which in turn provides a mirror like image of the original xyz.  

i notice that when i draw a line segment from x to x', that the points are parallel from each other -- they are exactly across from each other and do not intersect with any other points.  

yes, think i would see the same characteristic if you drew the line segment connecting y with the reflecting line and then y' with the reflecting line because of the essence of reflections. because it is a mirror image, each point was effected the same way which leads me to conclude that they same would apply to y and y'.

126 maybe. sorry