Math Problem Statement

Suatu fungsi pola kuadratik dengan informasi: - Bila x = 5 maka y = 15 - Bila x = 0 maka y = 20 - Bila x = 10 maka y = 0 Ditanyakan: a) Tentukan fungsinya b) Berapa nilai x agar y maksimum? c) Berapa nilai x agar x = y d) Gambar fungsinya

Solution

This problem asks for a quadratic function, given three points:

  • x=5x = 5, y=15y = 15
  • x=0x = 0, y=20y = 20
  • x=10x = 10, y=0y = 0

The general form of a quadratic function is:

y=ax2+bx+cy = ax^2 + bx + c

Let's solve step by step:

a) Determine the function

Substitute the points into the equation to form a system of equations.

  1. For x=5x = 5 and y=15y = 15: 25a+5b+c=1525a + 5b + c = 15

  2. For x=0x = 0 and y=20y = 20: c=20c = 20

  3. For x=10x = 10 and y=0y = 0: 100a+10b+c=0100a + 10b + c = 0

Since c=20c = 20, substitute this into the other two equations:

  1. 25a+5b+20=1525a+5b=525a + 5b + 20 = 15 \Rightarrow 25a + 5b = -5

  2. 100a+10b+20=0100a+10b=20100a + 10b + 20 = 0 \Rightarrow 100a + 10b = -20

Now, solve this system of equations:

  1. 25a+5b=525a + 5b = -5
  2. 100a+10b=20100a + 10b = -20

Divide the second equation by 2:

50a+5b=1050a + 5b = -10

Now subtract the first equation from this:

(50a+5b)(25a+5b)=10(5)(50a + 5b) - (25a + 5b) = -10 - (-5) 25a=525a = -5 a=15a = -\frac{1}{5}

Now substitute a=15a = -\frac{1}{5} into 25a+5b=525a + 5b = -5:

25(15)+5b=525\left(-\frac{1}{5}\right) + 5b = -5 5+5b=5-5 + 5b = -5 5b=0b=05b = 0 \Rightarrow b = 0

Thus, the quadratic function is:

y=15x2+20y = -\frac{1}{5}x^2 + 20

b) What is the value of xx when yy is maximum?

For a quadratic function y=ax2+bx+cy = ax^2 + bx + c, the vertex (maximum or minimum point) is given by x=b2ax = -\frac{b}{2a}.

Since b=0b = 0, the vertex occurs at x=0x = 0.

c) What is the value of xx when x=yx = y?

Set y=xy = x in the quadratic equation:

x=15x2+20x = -\frac{1}{5}x^2 + 20

Multiply everything by 5 to eliminate the fraction:

5x=x2+1005x = -x^2 + 100 x2+5x100=0x^2 + 5x - 100 = 0

Solve this quadratic equation using the quadratic formula:

x=5±524(1)(100)2(1)x = \frac{-5 \pm \sqrt{5^2 - 4(1)(-100)}}{2(1)} x=5±25+4002x = \frac{-5 \pm \sqrt{25 + 400}}{2} x=5±4252x = \frac{-5 \pm \sqrt{425}}{2} x=5±20.622x = \frac{-5 \pm 20.62}{2}

So, the solutions are:

x1=5+20.622=7.81andx2=520.622=12.81x_1 = \frac{-5 + 20.62}{2} = 7.81 \quad \text{and} \quad x_2 = \frac{-5 - 20.62}{2} = -12.81

d) Graph the function

The quadratic function is y=15x2+20y = -\frac{1}{5}x^2 + 20. It is an upside-down parabola with a maximum at (0,20)(0, 20) and intersects the x-axis at approximately x=10x = -10 and x=10x = 10.

Would you like a detailed plot of this function or further clarification?


Here are 5 additional related questions:

  1. How does changing the coefficient aa in a quadratic function affect its graph?
  2. How would you find the roots of a quadratic function using factoring instead of the quadratic formula?
  3. What is the significance of the vertex in a quadratic function?
  4. How do you calculate the discriminant of a quadratic equation, and what does it tell you about the solutions?
  5. What happens to the quadratic function if the constant cc is increased or decreased?

Tip: Always check whether the parabola opens upwards or downwards by inspecting the sign of the coefficient aa. A negative aa indicates it opens downwards (maximum point), and a positive aa means it opens upwards (minimum point).

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Math Problem Analysis

Mathematical Concepts

Algebra
Quadratic Functions
Systems of Equations

Formulas

Quadratic function y = ax^2 + bx + c
Vertex formula x = -b / 2a
Quadratic formula x = (-b ± √(b^2 - 4ac)) / 2a

Theorems

Vertex theorem for quadratic functions
Quadratic formula

Suitable Grade Level

Grade 9-11