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ValentinkaMS [17]
3 years ago
9

What is the vertex and the axis of symmetry of the quadratic function graphed below?

Mathematics
2 answers:
Kazeer [188]3 years ago
8 0
The answer is B ! as vertex is (6,3) and axis of symmetry lies on x=6 !

if you need explanation, comment !
Ede4ka [16]3 years ago
8 0

Answer:  The correct option is

(B)  The vertex is (6, 3), and the axis of symmetry is x = 6.

Step-by-step explanation:  We are given to find the vertex and the axis of symmetry of the graphed quadratic equation.

From the graph, we see that the graphed quadratic function is a parabola, which open downwards and maximum point is (6, 3).

So, the vertex of the parabola is at (6, 3).

Now, the axis of symmetry will be a line passing through the vertex and parallel to the Y-axis.

That is, x = k.

Since the axis of symmetry passes through the vertex (6, 3), so we get

x = 6.

Thus, the vertex is (6, 3), and the axis of symmetry is x = 6.

Option (B) is CORRECT.

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Dec 4,
sasho [114]

Answer:

y = -0.8x - 5

Step-by-step explanation:

x1 = (XA + XB) / 2 = -5

y1 = (YA + YB) / 2 = -1

m = (XA - XB) / (YA - YB) = -0

__________________________________________

(y - -1) = -0.8(x - -5)

y - -1 = -0.8x - 4

y = -0.8x - 4 + -1

y = -0.8x - 5

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3 years ago
a jet engine has a decibel level of 140dB, while a whisper has a decibel level of 30dB. How many times as intense as the sound o
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Answer:

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6 0
2 years ago
A train traveled the first 200 miles of its trip at a speed of 50 mph and the next 130 miles at a speed of 65 mph. What was the
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4 0
3 years ago
Help I gotta hand my homework in tomorrow
zvonat [6]
Simple...

3*10=5*6

2*8=4*4

21+11=9+23

23+20=2+41

\frac{36}{9} = \frac{12}{3}

30-5=40-15

90-80=20-10

Thus, your answers.
6 0
3 years ago
Read 2 more answers
A has the coordinates (–4, 3) and B has the coordinates (4, 4). If DO,1/2(x, y) is a dilation of △ABC, what is true about the im
loris [4]

Answer:

Coordinates of Vertices of triangle ABC are A (-4,3) , B(4,4) and C(1,1).

As, DO is Dilation of Δ ABC by Scale factor of \frac{1}{2}.

Vertices of A' B'C' are

A'=(\frac{-4}{2},\frac{3}{2})=(-2,\frac{3}{2}), B'=(\frac{4}{2},\frac{4}{2})=(2,2), C'=(\frac{1}{2},\frac{1}{2})

So, Image Δ A'B'C' will be smaller than the Pre image Δ ABC.

The two triangles will be congruent.

AO is Dilated by a factor of half , so A'O' will be half of AO.

So, correct Statements are

1. AB is parallel to A'B'.

2.DO,1/2(x, y) =  

The distance from A' to the origin is half the distance from A to the origin.

OA=\sqrt{(-4)^2+3^2}=\sqrt{25}=5\\\\ O'A'=\sqrt{2^2+(\frac{3}{2})^2}=\frac{5}{2}=2.5

8 0
3 years ago
Read 2 more answers
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