Resources

Resources

Bytelearn - cat image with glasses

AI tutor

Welcome to Bytelearn!

Let’s check out your problem:

f(x)=(x+6)^(2)-16
At what values of x does the graph of the function intersect the x-axis?
Choose 1 answer:
(A) x=-2,x=10
(B) x=2,x=-10
(C) x=-2,x=-10
(D) f(x) does not intersect the x-axis.

$f(x)=(x+6)^2-16$ $\newline$ At what values of $x$ does the graph of the function intersect the $x$ -axis? $\newline$ Choose $1$ answer: $\newline$ (A) $x=-2$ , $x=10$ $\newline$ (B) $x=2$ , $x=-10$ $\newline$ (C) $x=-2$ , $x=-10$ $\newline$ (D) $f(x)$ does not intersect the $x$ -axis.

View step-by-step help

View step-by-step help

Reflections of functions

Full solution

Q.

f(x)=(x+6)^2-16

\newline

At what values of

x

does the graph of the function intersect the

x

-axis?

\newline

Choose

1

answer:

\newline

(A)

x=-2

,

x=10

\newline

(B)

x=2

,

x=-10

\newline

(C)

x=-2

,

x=-10

\newline

(D)

f(x)

does not intersect the

x

-axis.

Set Equation and Solve: To find the $x$ -intercepts of the graph of the function, we need to set $f(x)$ to $0$ and solve for $x$ . So, we set the equation $(x+6)^2 - 16 = 0$ .
Solve for x: Now we solve the equation for x. $\newline$ $(x+6)^2 - 16 = 0$ $\newline$ $(x+6)^2 = 16$
Take Square Root: Next, we take the square root of both sides of the equation. $\newline$ $\sqrt{(x+6)^2} = \pm\sqrt{16}$ $\newline$ $x + 6 = \pm4$
Two Equations to Solve: We now have two equations to solve for $x$ : $1.$ $x + 6 = 4$ $2.$ $x + 6 = -4$
Solve for x (Equation $1$ ): Solving the first equation for x: $\newline$ $x + 6 = 4$ $\newline$ $x = 4 - 6$ $\newline$ $x = -2$
Solve for x (Equation $2$ ): Solving the second equation for x: $\newline$ $x + 6 = -4$ $\newline$ $x = -4 - 6$ $\newline$ $x = -10$

More problems from Reflections of functions

Question

What does the transformation

f(x) \mapsto f(x) - 6

do to the graph of

f(x)

\newline

\newline

\text{translates it } 6 \text{ units down}

\newline

\text{translates it } 6 \text{ units right}

\newline

\text{translates it } 6 \text{ units left}

\newline

\text{translates it } 6 \text{ units up}

Posted 3 months ago

Question

g(x)

g(x)

is the reflection across the

y

f(x) = -4x + 1

\newline

\newline

\text{[g(x) = 4x - 1]}

\newline

\text{[g(x) = -4x + 1]}

\newline

\text{[g(x) = 4x + 1]}

\newline

\text{[g(x) = -4x - 1]}

Posted 3 months ago

Question

What kind of transformation converts the graph of

f(x) = -8(x - 4)^2 + 6

into the graph of

g(x) = -8(x + 6)^2 + 6

\newline

\newline

\text{[A]translation 10 units left}

\newline

\text{[B]translation 10 units right}

\newline

\text{[C]translation 10 units up}

\newline

\text{[D]translation 10 units down}

Posted 3 months ago

Question

g(x)

g(x)

is the reflection across the x-axis of

f(x) = -6(x - 7)^2 + 2

\newline

\newline

g(x) = 6(x - 7)^2 - 2

\newline

g(x) = -6(x + 7)^2 + 2

\newline

g(x) = -6(x - 7)^2 - 2

\newline

g(x) = 6(x + 7)^2 - 2

Posted 3 months ago

Question

\lim _{x \rightarrow \frac{\pi}{4}} \cos (x)=?

\newline

1

\newline

\frac{1}{2}

\newline

1

\newline

\frac{\sqrt{2}}{2}

\newline

(D) The limit doesn't exist.

Posted 3 months ago

Question

Tess tried to solve the differential equation

\frac{d y}{d x}=2 x y-6 x

. This is her work:

\newline

\frac{d y}{d x}=2 x y-6 x

\newline

1

\quad \frac{d y}{d x}=(y-3) 2 x

\newline

2

\quad \int \frac{1}{y-3} d y=\int 2 x d x

\newline

3

\quad \ln |y-3|=x^{2}+C_{1}

\newline

4

\quad e^{\ln |y-3|}=e^{x^{2}}+C_{1}

\newline

5

\quad|y-3|=e^{x^{2}}+C_{1}

\newline

6

\quad y-3= \pm e^{x^{2}}+C_{2}

\newline

7

\quad y= \pm e^{x^{2}}+C

\newline

Is Tess's work correct? If not, what is her mistake?

\newline

1

\newline

(A) Tess's work is correct.

\newline

1

is incorrect. We're not allowed to factor an expression when solving differential equations.

\newline

4

is incorrect. The right-hand side of the equation should be

e^{x^{2}+C_{1}}

\newline

7

is incorrect. Tess didn't account for adding

3

to the right side.

Posted 2 months ago

Question

Aditya tried to solve the differential equation

\frac{d y}{d x}=6 x-30

. This is his work:

\newline

\frac{d y}{d x}=6 x-30

\newline

1

\quad \int 30 d y=\int 6 x d x

\newline

2

\quad 30 y=3 x^{2}+C

\newline

3

\quad y=\frac{x^{2}}{10}+C

\newline

Is Aditya's work correct? If not, what is his mistake?

\newline

1

\newline

(A) Aditya's work is correct.

\newline

1

is incorrect. The separation of variables wasn't done correctly.

\newline

2

is incorrect. Aditya didn't integrate

30

\newline

3

is incorrect. The right-hand side of the equation should be

\frac{30}{3 x^{2}+C}

Posted 3 months ago

Question

Jonael dropped a sandbag from a hot air balloon at a target that is

250

meters below the balloon. At this moment, the sandbag is

75

meters below the balloon.

\newline

2

of the following expressions represents the distance between the sandbag and the target?

\newline

2

\newline

|-250+75|

\newline

|250+75|

\newline

|-75+250|

Posted 2 months ago

Question

f(x) = -(x+1)(x+7)

\newline

The function represents a parabola in the

xy

-plane. Which of the following is an equivalent form of

f

y

-intercept of the graph of

f

appears as a constant or coefficient?

\newline

1

\newline

f(x) = -(x+4)^{2}+9

\newline

f(x) = -(x+4)^{2}-9

\newline

f(x) = -x^{2}+8x+7

\newline

f(x) = -x^{2}-8x-7

Posted 2 months ago

Question

Determine whether the function

f(x)

is continuous at

x=-3

\newline

f(x)=\left\{\begin{array}{ll} 18-x^{2}, & x \leq-3 \\ 15+3 x, & x>-3 \end{array}\right.

\newline

f(x)

is discontinuous at

x=-3

\newline

f(x)

is continuous at

x=-3

Posted 3 months ago