Find the equation of the axis of symmetry and the coordinates of the vertex of the graph of the function y = 2x2 + 4x –3

Mathematics

2 answers:

egoroff_w [7]3 years ago

8 0

Your vertex would be (-1,-5)
and your axis of symmetry is x=-1

Oksana_A [137]3 years ago

7 0

Answer:

Axis of symmetry is x=-1.

The vertex of the graph is (-1,-5).

Step-by-step explanation:

If a quadratic function is $f(x)=ax^2=bx+c$ , then the axis of symmetry is

$x=-\frac{b}{2a}$

$Vertex=(-\frac{b}{2a},f(-\frac{b}{2a}))$

The given equation is

$y=2x^2+4x-3$

Here,

$a=2,b=4,c=-3$

The axis of symmetry is

$x=-\frac{4}{2(2)}=-1$

Substitute x=-1 in the given equation to find the y-coordinate of the vertex.

$y=2(-1)^2+4(-1)-3$

$y=2-4-3$

$y=-5$

Therefore, the vertex of the graph is (-1,-5).

You might be interested in

Rectangle and a square have the same area. the length of the rectangle is seventy feet more than two times its width. the length

Natasha2012 [34]

LxW=D
length
width
dimensions
it might be backwards two <span />

6 0

3 years ago

"When in the Course of human events, it becomes necessary for one people to dissolve the political bands which have connected th

Vesna [10]

When within the Course of human occasions, it gets to be fundamental for individuals to break up the political groups which have associated them with another, and to look forward to the powers of this world, the partitioned and the same to station to which the Laws of Nature and of Nature's God entitle them, a conventional regard to the conclusions of mankind requires that they ought to pronounce the causes which.

3 0

2 years ago

1. A sine function has the following key features:

andrew11 [14]

Problem 1

See the attached image (figure 1)

16pi seems like a typo. I'm going to assume that it's a fraction and it is 1/(6pi)
f = 1/(6pi) = frequency
T = 1/f = 1/(1/(6pi)) = 6pi
Amplitude = 2
a = 2
b = 2pi/T = 2pi/(6pi) = 1/3
Midline: y = 3
d = 3

The function is
y = a*sin(bx-c)+d
y = 2*sin(1/3*x-0)+3
y = 2*sin(x/3)+3

===============================================

Problem 2

See the attached image (figure 2)

T = 12 is the period
a = 4 is the amplitude
b = 2pi/T = 2pi/12 = pi/6
y = 1 is the midline so d = 1
The y intercept is (0,1) which is the midline, which indicates no phase shifts have occurred so c = 0

The function is
y = a*sin(bx-c)+d
y = 4*sin((pi/6)x-0)+1
y = 4*sin((pi/6)x)+1

===============================================

Problem 3

See the attached image (figure 3)

Period = 4pi
T = 4pi
b = 2pi/T = 2pi/(4pi) = 1/2 = 0.5
Amplitude = 2
a = 2
Midline: y = 3
d = 3
y-intercept: (0,3)
The function is a reflection of its parent function over the x-axis, so 'a' is negative meaning a = -2 instead of a = 2

The function is
y = a*sin(bx-c)+d
y = -2*sin(0.5x-0)+3
y = -2*sin(0.5x)+3

===============================================

Problem 4

See the attached image (figure 4)

a = 10 which is half of the distance between the highest and lowest points
T = 8 is the period
b = 2pi/T = 2pi/8 = pi/4
c = -pi/2 is the phase shift since its really a cosine graph
d = 0 is the midline

The function is
y = a*sin(bx-c)+d
y = 10*sin((pi/4)*x+(-pi/2))+0
y = 10*sin((pi/4)*x+pi/2)

===============================================

Problem 5

See the attached image (figure 5)

a = 2 is the amplitude since it bobs up and down this distance from the midline
T = 8 seconds is the period (double that of the time it takes for it to go from the highest to the lowest point)
b = 2pi/T = 2pi/8 = pi/4
c = 0 is the phase shift as the buoy starts at normal depth of 20 meters
d = 20 is the midline

The function is
y = a*sin(bx-c)+d
y = 2*sin((pi/4)x-0)+20
y = 2*sin((pi/4)x)+20

===============================================