The graph of f(x) = 4/(x+2) + 3 is drawn below. C is the point of intersection of the asymptotes of f and B…

Question

The graph of f(x) = 4/(x+2) + 3 is drawn below. C is the point of intersection of the asymptotes of f and B is the x-intercept of f.

ScanSolve AI Tutor · Accepted Answer

Here's the solution to Question 6: The given function is f(x) = (4)/(x+2) + 3. 6.1 Write down the equation of the vertical asymptote of f. Step 1: The vertical asymptote occurs where the denominator of the rational part of the function is zero. x+2 = 0; x = −2; The equation of the vertical asymptote is x=−2. 6.2 Write down the range of f. Step 1: The horizontal asymptote of the function f(x) = (a)/(x−p) + q is y=q. In this case, q=3. The range of a hyperbola is all real numbers except the value of its horizontal asymptote. The range of f is y ∈ R, y ≠ 3. 6.3 Calculate the coordinates of B. Step 1: B is the x-intercept of f. To find the x-intercept, set f(x)=0. 0 = (4)/(x+2) + 3 Step 2: Solve for x. −3 = (4)/(x+2); −3(x+2) = 4; −3x − 6 = 4; −3x = 10; x = −10/3 The coordinates of B are (−10/3, 0). 6.4 Determine the values of k for which y=−4x+k is a tangent to f. Step 1: Set the function f(x) equal to the line y=−4x+k. (4)/(x+2) + 3 = −4x+k Step 2: Multiply both sides by (x+2) to eliminate the denominator. 4 + 3(x+2) = (−4x+k)(x+2); 4 + 3x + 6 = −4x² − 8x + kx + 2k; 3x + 10 = −4x² + (k−8)x + 2k; Step 3: Rearrange the equation into the standard quadratic form ax²+bx+c=0. 4x² + 3x − (k−8)x + 10 − 2k = 0; 4x² + (3 − k + 8)x + (10 − 2k) = 0; 4x² + (11 − k)x + (10 − 2k) = 0; Step 4: For the line to be a tangent, there must be exactly one point of intersection, which means the discriminant (Δ) of the quadratic equation must be zero. Δ = b² − 4ac = 0. Here, a=4, b=(11−k), c=(10−2k). (11−k)² − 4(4)(10−2k) = 0; (121 − 22k + k²) − 16(10−2k) = 0; 121 − 22k + k² − 160 + 32k = 0; k² + 10k − 39 = 0 Step 5: Solve the quadratic equation for k. (k+13)(k−3) = 0; k+13=0 or k−3=0; k=−13 or k=3; The values of k are k=−13 or k=3. 6.5 Calculate the distance between B and D, where D is the reflection of B in the line of symmetry of f, having a negative gradient. Step 1: Identify the coordinates of B and the equation of the line of symmetry with a negative gradient. From 6.3, B is (−10/3, 0). The point of intersection of the asymptotes (C) is (−2, 3). The lines of symmetry pass through C. The general equations for the lines of symmetry of f(x) = (a)/(x−p) + q are y−q = ±(x−p). So, y−3 = ±(x−(−2)); y−3 = ±(x+2); The line of symmetry with a negative gradient is y−3 = −(x+2). y = −x−2+3; y = −x+1; Step 2: Find the coordinates of D, the reflection of B in the line y=−x+1. Let B be (x_B, y_B) = (−10/3, 0) and D be (x_D, y_D). The line segment BD is perpendicular to the line of symmetry y=−x+1. The gradient of the line of symmetry is −1, so the gradient of BD is 1. The equation of the line passing through B and D is y − y_B = 1(x − x_B). y − 0 = 1(x − (−10/3)); y = x + 10/3 Step 3: Find the midpoint M of BD, which is the intersection of the line BD and the line of symmetry. x + 10/3 = −x+1; 2x = 1 − 10/3; 2x = (3−10)/(3); 2x = −7/3; x_M = −7/6 Substitute x_M into the line of symmetry equation to find y_M: y_M = −(−7/6) + 1; y_M = 7/6 + 6/6; y_M = 13/6 So, the midpoint M is (−7/6, 13/6). Step 4: Use the midpoint formula to find the coordinates of D. x_M = (x_B+x_D)/(2) ⇒ −7/6 = (−10/3+x_D)/(2); −7/3 = −10/3+x_D; x_D = −7/3 + 10/3 = 3/3 = 1; y_M = (y_B+y_D)/(2) ⇒ 13/6 = (0+y_D)/(2); y_D = 13/3 So, D is (1, 13/3). Step 5: Calculate the distance between B(−10/3, 0) and D(1, 13/3) using the distance formula d = √((x₂−x₁)² + (y₂−y₁)²). dBD = √((1 − (−10/3))² + (13/3 − 0)²); dBD = √((3/3 + 10/3)² + (13/3)²); dBD = √((13/3)² + (13/3)²); dBD = √(2 · (13/3)²); dBD = 13/3√(2) The distance between B and D is (13√(2))/(3).

Point	x	y
x-intercept	−3.33	0
y-intercept	0	5

The graph of f(x) = 4/(x+2) + 3 is drawn below. C is the point of intersection of the asymptotes of f and B is the x-intercept of f.

Answers

Step-by-step solution

Part 6.1· 1 step

The vertical asymptote occurs where the denominator of the rational part of the function is zero.

Part 6.2· 1 step
$y \in \mathbb{R}, y \neq 3$

The horizontal asymptote of the function $f(x) = \frac{a}{x-p} + q$ is $y=q$ .

Part 6.3· 2 steps
$\left(-\frac{10}{3}, 0\right)$

B is the x-intercept of $f$ .

Solve for $x$ .

Part 6.4· 5 steps
$k=-13 \text{ or } k=3$

Set the function $f(x)$ equal to the line $y=-4x+k$ .

Multiply both sides by $(x+2)$ to eliminate the denominator.

Rearrange the equation into the standard quadratic form $ax^2+bx+c=0$ .

For the line to be a tangent, there must be exactly one point of intersection, which means the discriminant ( $\Delta$ ) of the quadratic equation must be zero.

Solve the quadratic equation for $k$ .

Part 6.5· 5 steps
$\frac{13\sqrt{2}}{3}$

Identify the coordinates of B and the equation of the line of symmetry with a negative gradient.

Find the coordinates of D, the reflection of B in the line $y=-x+1$ .

Find the midpoint M of BD, which is the intersection of the line BD and the line of symmetry.

Use the midpoint formula to find the coordinates of D.

Calculate the distance between B $\left(-\frac{10}{3}, 0\right)$ and D $\left(1, \frac{13}{3}\right)$ using the distance formula $d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$ .

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The graph of f(x) = 4/(x+2) + 3 is drawn below. C is the point of intersection of the asymptotes of f and B is the x-intercept of f.

Answers

Step-by-step solution

Part 6.1· 1 step

The vertical asymptote occurs where the denominator of the rational part of the function is zero.

Part 6.2· 1 stepy∈R,y≠3y \in \mathbb{R}, y \neq 3y∈R,y=3

The horizontal asymptote of the function f(x)=ax−p+qf(x) = \frac{a}{x-p} + qf(x)=x−pa​+q is y=qy=qy=q.

Part 6.3· 2 steps(−103,0)\left(-\frac{10}{3}, 0\right)(−310​,0)

B is the x-intercept of fff.

Solve for xxx.

Part 6.4· 5 stepsk=−13 or k=3k=-13 \text{ or } k=3k=−13 or k=3

Set the function f(x)f(x)f(x) equal to the line y=−4x+ky=-4x+ky=−4x+k.

Multiply both sides by (x+2)(x+2)(x+2) to eliminate the denominator.

Rearrange the equation into the standard quadratic form ax2+bx+c=0ax^2+bx+c=0ax2+bx+c=0.

For the line to be a tangent, there must be exactly one point of intersection, which means the discriminant (Δ\DeltaΔ) of the quadratic equation must be zero.

Solve the quadratic equation for kkk.

Part 6.5· 5 steps1323\frac{13\sqrt{2}}{3}3132​​

Identify the coordinates of B and the equation of the line of symmetry with a negative gradient.

Find the coordinates of D, the reflection of B in the line y=−x+1y=-x+1y=−x+1.

Find the midpoint M of BD, which is the intersection of the line BD and the line of symmetry.

Use the midpoint formula to find the coordinates of D.

Calculate the distance between B(−103,0)\left(-\frac{10}{3}, 0\right)(−310​,0) and D(1,133)\left(1, \frac{13}{3}\right)(1,313​) using the distance formula d=(x2−x1)2+(y2−y1)2d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}d=(x2​−x1​)2+(y2​−y1​)2​.

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Part 6.2· 1 step
$y \in \mathbb{R}, y \neq 3$

The horizontal asymptote of the function $f(x) = \frac{a}{x-p} + q$ is $y=q$ .

Part 6.3· 2 steps
$\left(-\frac{10}{3}, 0\right)$

B is the x-intercept of $f$ .

Solve for $x$ .

Part 6.4· 5 steps
$k=-13 \text{ or } k=3$

Set the function $f(x)$ equal to the line $y=-4x+k$ .

Multiply both sides by $(x+2)$ to eliminate the denominator.

Rearrange the equation into the standard quadratic form $ax^2+bx+c=0$ .

For the line to be a tangent, there must be exactly one point of intersection, which means the discriminant ( $\Delta$ ) of the quadratic equation must be zero.

Solve the quadratic equation for $k$ .

Part 6.5· 5 steps
$\frac{13\sqrt{2}}{3}$

Find the coordinates of D, the reflection of B in the line $y=-x+1$ .

Calculate the distance between B $\left(-\frac{10}{3}, 0\right)$ and D $\left(1, \frac{13}{3}\right)$ using the distance formula $d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$ .