Ofind the minimum value of the quadratic expression −4x2+8x−25,
−
4
x
2
+
8
x
−
25
,
marla used the following steps to complete the square:

step 1: −4(x2+8x)−25
−
4
(
x
2
+
8
x
)
−
25

step 2: −4(x2+8x+16−16)−25
−
4
(
x
2
+
8
x
+
16
−
16
)
−
25

step 3: −4(x2+8x+16)+64−25
−
4
(
x
2
+
8
x
+
16
)
+
64
−
25

step 4: −4(x+4)2+39
−
4
(
x
+
4
)
2
+
39

did marla use the correct steps to complete the square?

Marla didn't use the right steps to complete the square. Maria made a mistake in step 1, she put 8x instead of -2x

Step-by-step explanation:

we have

This is a vertical parabola open downward

The vertex is a maximum

Find the vertex

step 1

Factor the leading coefficient -4

step 2

Complete the square

step 3

step 4

Rewrite as perfect squares

the vertex is the point (1,-21)

so

The maximum value of the quadratic equation is (1,-21)

therefore

Marla didn't use the right steps to complete the square. Maria made a mistake in step 1, she put 8x instead of -2x

f(x) = - 2/3x + 5; f(5/2)

substitute 5/2 in for x:

f(5/2) = -2/3(5/2) + 5

simplify the right side:

f(5/2) = -5/3 + 5

f(5/2) = 10/3

answer:

f(5/2) = 10/3

i hope this you!

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'.