The graphs of f(x) = ax2 + bx + c and g(x) = -4/(x-3) + 8 are drawn below. P is the turning point of f and the point of intersection of the asymptotes of g. M and T are x-intercepts of f and g respectively.

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Asked on August 11, 2026Mathematics

This mathematics problem involves applying core mathematical principles and formulas. Below you will find a complete step-by-step solution with detailed explanations for each step, helping you understand not just the answer but the method behind it.

Accepted Answer · 2026-08-11T15:34:26.476Z

Step 1: Determine the domain of g(x). The function g(x) = (-4)/(x-3) + 8 is a rational function. The denominator cannot be zero. x-3 ≠ 0 x ≠ 3 The domain of g is all real numbers except x=3. The domain of g is x R, x ≠ 3. Step 2: Determine the range of f(x). The function f(x) is a parabola with turning point P (3; 8). From the graph, the parabola opens downwards, meaning the turning point is a maximum. Therefore, the maximum value of f(x) is the y-coordinate of the turning point. The range of f is y ≤ 8. Step 3: Determine the values of x for which g(x) ≤ f(x). From the graph, the function g(x) is below or equal to f(x) between the vertical asymptote x=3 and the intersection point D (5; 6). Since x=3 is an asymptote for g(x), x cannot be equal to 3. At point D, x=5, g(5)=f(5)=6, so x=5 is included. The values of x for which g(x) ≤ f(x) are 3 < x ≤ 5. Step 4: Determine the values of x for which f(x) < 6. First, we need the equation of f(x). The turning point is P (3; 8), so f(x) = a(x-3)^2 + 8. The graph passes through D (5; 6). Substitute D into the equation: 6 = a(5-3)^2 + 8 6 = a(2)^2 + 8 6 = 4a + 8 4a = -2 a = -(1)/(2) So, f(x) = -(1)/(2)(x-3)^2 + 8. Now, we solve f(x) < 6: -(1)/(2)(x-3)^2 + 8 < 6 -(1)/(2)(x-3)^2 < -2 Multiply by -2 and reverse the inequality sign: (x-3)^2 > 4 Take the square root of both sides: sqrt((x-3)^2) > sqrt(4) |x-3| > 2 This implies two separate inequalities: x-3 > 2 or x-3 < -2 x > 5 or x < 1 The values of x for which f(x) < 6 are x < 1 or x > 5. Step 5: Show that the equation of the parabola is f(x) = -(1)/(2)x^2 + 3x + (7)/(2). From Step 4, we found f(x) = -(1)/(2)(x-3)^2 + 8. Expand the expression: f(x) = -(1)/(2)(x^2 - 6x + 9) + 8 f(x) = -(1)/(2)x^2 + 3x - (9)/(2) + 8 f(x) = -(1)/(2)x^2 + 3x - (9)/(2) + (16)/(2) f(x) = -(1)/(2)x^2 + 3x + (7)/(2) The equation is shown to be f(x) = -(1)/(2)x^2 + 3x + (7)/(2). Step 6: Calculate the length of MT. M is the x-intercept of f(x), and T is the x-intercept of g(x). To find T, set g(x) = 0: 0 = (-4)/(x-3) + 8 -8 = (-4)/(x-3) -8(x-3) = -4 2(x-3) = 1 2x - 6 = 1 2x = 7 x = (7)/(2) So, T is ((7)/(2); 0). To find M, set f(x) = 0: 0 = -(1)/(2)x^2 + 3x + (7)/(2) Multiply by 2 to clear fractions: 0 = -x^2 + 6x + 7 x^2 - 6x - 7 = 0 Factor the quadratic equation: (x-7)(x+1) = 0 x = 7 or x = -1 From the graph, M is the negative x-intercept, so M is (-1; 0). The length of MT is the distance between the x-coordinates of M and T: Length MT = |(7)/(2) - (-1)| Length MT = |(7)/(2) + 1| Length MT = |(7)/(2) + (2)/(2)| Length MT = |(9)/(2)| The length of MT is (9)/(2) units. Step 7: Determine the equation of the tangent to f at D. The point D is (5; 6). To find the equation of the tangent line, we need its slope, which is the derivative of f(x) evaluated at x=5. f(x) = -(1)/(2)x^2 + 3x + (7)/(2) Find the derivative f'(x): f'(x) = (d)/(dx)(-(1)/(2)x^2 + 3x + (7)/(2)) f'(x) = -x + 3 Now, evaluate f'(x) at x=5 to find the slope m: m = f'(5) = -5 + 3 m = -2 Using the point-slope form of a linear equation, y - y_1 = m(x - x_1), with D (5; 6) and m = -2: y - 6 = -2(x - 5) y - 6 = -2x + 10 y = -2x + 10 + 6 y = -2x + 16 The equation of the tangent to f at D is y = -2x + 16. Send me the next one 📸

Accepted Answer · 2026-08-11T15:34:26.476Z

Step 1: Determine the domain of g(x). The function g(x) = (-4)/(x-3) + 8 is a rational function. The denominator cannot be zero. x-3 ≠ 0 x ≠ 3 The domain of g is all real numbers except x=3. The domain of g is x R, x ≠ 3. Step 2: Determine the range of f(x). The function f(x) is a parabola with turning point P (3; 8). From the graph, the parabola opens downwards, meaning the turning point is a maximum. Therefore, the maximum value of f(x) is the y-coordinate of the turning point. The range of f is y ≤ 8. Step 3: Determine the values of x for which g(x) ≤ f(x). From the graph, the function g(x) is below or equal to f(x) between the vertical asymptote x=3 and the intersection point D (5; 6). Since x=3 is an asymptote for g(x), x cannot be equal to 3. At point D, x=5, g(5)=f(5)=6, so x=5 is included. The values of x for which g(x) ≤ f(x) are 3 < x ≤ 5. Step 4: Determine the values of x for which f(x) < 6. First, we need the equation of f(x). The turning point is P (3; 8), so f(x) = a(x-3)^2 + 8. The graph passes through D (5; 6). Substitute D into the equation: 6 = a(5-3)^2 + 8 6 = a(2)^2 + 8 6 = 4a + 8 4a = -2 a = -(1)/(2) So, f(x) = -(1)/(2)(x-3)^2 + 8. Now, we solve f(x) < 6: -(1)/(2)(x-3)^2 + 8 < 6 -(1)/(2)(x-3)^2 < -2 Multiply by -2 and reverse the inequality sign: (x-3)^2 > 4 Take the square root of both sides: sqrt((x-3)^2) > sqrt(4) |x-3| > 2 This implies two separate inequalities: x-3 > 2 or x-3 < -2 x > 5 or x < 1 The values of x for which f(x) < 6 are x < 1 or x > 5. Step 5: Show that the equation of the parabola is f(x) = -(1)/(2)x^2 + 3x + (7)/(2). From Step 4, we found f(x) = -(1)/(2)(x-3)^2 + 8. Expand the expression: f(x) = -(1)/(2)(x^2 - 6x + 9) + 8 f(x) = -(1)/(2)x^2 + 3x - (9)/(2) + 8 f(x) = -(1)/(2)x^2 + 3x - (9)/(2) + (16)/(2) f(x) = -(1)/(2)x^2 + 3x + (7)/(2) The equation is shown to be f(x) = -(1)/(2)x^2 + 3x + (7)/(2). Step 6: Calculate the length of MT. M is the x-intercept of f(x), and T is the x-intercept of g(x). To find T, set g(x) = 0: 0 = (-4)/(x-3) + 8 -8 = (-4)/(x-3) -8(x-3) = -4 2(x-3) = 1 2x - 6 = 1 2x = 7 x = (7)/(2) So, T is ((7)/(2); 0). To find M, set f(x) = 0: 0 = -(1)/(2)x^2 + 3x + (7)/(2) Multiply by 2 to clear fractions: 0 = -x^2 + 6x + 7 x^2 - 6x - 7 = 0 Factor the quadratic equation: (x-7)(x+1) = 0 x = 7 or x = -1 From the graph, M is the negative x-intercept, so M is (-1; 0). The length of MT is the distance between the x-coordinates of M and T: Length MT = |(7)/(2) - (-1)| Length MT = |(7)/(2) + 1| Length MT = |(7)/(2) + (2)/(2)| Length MT = |(9)/(2)| The length of MT is (9)/(2) units. Step 7: Determine the equation of the tangent to f at D. The point D is (5; 6). To find the equation of the tangent line, we need its slope, which is the derivative of f(x) evaluated at x=5. f(x) = -(1)/(2)x^2 + 3x + (7)/(2) Find the derivative f'(x): f'(x) = (d)/(dx)(-(1)/(2)x^2 + 3x + (7)/(2)) f'(x) = -x + 3 Now, evaluate f'(x) at x=5 to find the slope m: m = f'(5) = -5 + 3 m = -2 Using the point-slope form of a linear equation, y - y_1 = m(x - x_1), with D (5; 6) and m = -2: y - 6 = -2(x - 5) y - 6 = -2x + 10 y = -2x + 10 + 6 y = -2x + 16 The equation of the tangent to f at D is y = -2x + 16. Send me the next one 📸

The graphs of f(x) = ax2 + bx + c and g(x) = -4/(x-3) + 8 are drawn below. P is the turning point of f and the point of intersection of the asymptotes of g. M and T are x-intercepts of f and g respectively.

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The graphs of f(x) = ax2 + bx + c and g(x) = -4/(x-3) + 8 are drawn below. P is the turning point of f and the point of intersection of the asymptotes of g. M and T are x-intercepts of f and g respectively.

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