All categories

3 years ago

What is the vertex of the graph of the equation y= 3x^2+6x+1

1 answer:

8 0

$y=a(x-h)^2+k \Rightarrow \hbox{vertex}=(h,k)\\y= 3x^2+6x+1\\ y=3x^2+6x+3-2\\ y=3(x^2+2x+1)-2\\ y=3(x+1)^2-2\Rightarrow h=-1,k=-2\\\\ (-1,-2)$

You might be interested in

Order these numbers from least to greatest:<br> 12 1.1 53 2 0 0.4 −2 58

wel

Answer:

-2, 0, 0.4, 1.1, 2, 12, 53, 58

Step-by-step explanation:

7 0

4 years ago

Read 2 more answers

YALL NEVER ANSWER AH POOP plz help

lakkis [162]

Answer:

The given point A (6,13) lies on the equation. True

The given point B(21,33) lies on the equation. True

The given point C (99, 137) lies on the equation. True

Step-by-step explanation:

Here, the given equation is : $y =\frac{4}{3} x + 5$

Now,check the given equation for the given points.

1) A (6,13)

Substitute x = 6 in the given equation $y =\frac{4}{3} x + 5$

$y =\frac{4}{3} (6) + 5 = 4(2) + 5 = 8 + 5 = 13$

⇒ y= 13

Hence, the given point A (6,13) lies on the equation.

2) B (21,33)

Substitute x = 21 in the given equation $y =\frac{4}{3} x + 5$

$y =\frac{4}{3} (21) + 5 = 4(7) + 5 = 28 + 5 = 33$

⇒ y = 33

Hence, the given point B(21,33) lies on the equation.

3) C (99, 137)

Substitute x = 99 in the given equation $y =\frac{4}{3} x + 5$

$y =\frac{4}{3} (99) + 5 = 4(33) + 5 = 132 + 5 = 137$

⇒ y = 137

Hence, the given point C (99, 137) lies on the equation.

7 0

3 years ago

Read 2 more answers

Lesson: 1.08Given this function: f(x) = 4 cos(TTX) + 1Find the following and be sure to show work for period, maximum, and minim

Ber [7]

The given function is

$f(x)=4\cos \text{(}\pi x)+1$

The general form of the cosine function is

$y=a\cos (bx+c)+d$

a is the amplitude

2pi/b is the period

c is the phase shift

d is the vertical shift

By comparing the two functions

a = 4

b = pi

c = 0

d = 1

Then its period is

$\begin{gathered} \text{Period}=\frac{2\pi}{\pi} \\ \text{Period}=2 \end{gathered}$

The equation of the midline is

$y_{ml}=\frac{y_{\max }+y_{\min }}{2}$

Since the maximum is at the greatest value of cos, which is 1, then

$\begin{gathered} y_{\max }=4(1)+1 \\ y_{\max }=5 \end{gathered}$

Since the minimum is at the smallest value of cos, which is -1, then

$\begin{gathered} y_{\min }=4(-1)+1 \\ y_{\min }=-4+1 \\ y_{\min }=-3 \end{gathered}$

Then substitute them in the equation of the midline

$\begin{gathered} y_{ml}=\frac{5+(-3)}{2} \\ y_{ml}=\frac{2}{2} \\ y_{ml}=1 \end{gathered}$

The answers are:

Period = 2

Equation of the midline is y = 1

Maximum = 5

Minimum = -3

3 0

1 year ago

Solve the system of linear equations. <br>- 3x + 5y = 13 <br> x + 4y = -10

Molodets [167]

Answer:

(-6,-1)

Step-by-step explanation:

5 0

3 years ago

DEFG models the shape of a pig pen on a farm. What is the perimeter of the pen?

Vanyuwa [196]

Umm I though it was 14

6 0

3 years ago