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Practice Problems Simplified Revision Notes

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Practice Problems

Problems:


Problem 1: Evaluating a Function

infoNote

Question: Given the function f(x)=2x23x+4f(x) = 2x^2 - 3x + 4, find:

  1. f(1)f(1)
  2. f(2)f(-2)

Explanation**:**

When you see a function like this, it's like a machine that takes an input number (in this case, xx), does some calculations, and gives you an output. Here, you're asked to find out what happens when you put x=1x = 1 and x=2x = -2 into the machine.


Problem 2: Finding Inputs for a Function

infoNote

Question: If f(x)=5x8f(x) = 5x - 8, find the value of xx when f(x)=17f(x) = 17.

Explanation**:**

Normally, you would put a number into a function and find the result. Here, you're given the result (17)(17) and need to figure out what number you started with (that is, xx). To do this, you'll work backward, like solving a puzzle in reverse.


Problem 3: Graphing a Linear Function

infoNote

Question: Graph the linear function y=2x+3y = -2x + 3 for xx values from2 -2 to 22.

Explanation**:**

To draw a picture of this function on a graph, you need to find out what yy is for different values of xx. After you have a few points, you can connect them to see the full picture, which will be a straight line.


Problem 4: Finding the Roots of a Quadratic Function

infoNote

Question: Find the roots of the quadratic function f(x)=x25x+6f(x) = x^2 - 5x + 6.

Explanation**:**

The "roots" of a function are the points where the function touches the x-axis (where y=0y = 0). To find these points, we solve the equation by setting f(x)=0f(x) = 0. You can find these points by factorizing the equation or using a special formula called the quadratic formula.


Problem 5: Solving a System of Linear Functions

infoNote

Question: If f(x)=3x+2f(x) = 3x + 2 and g(x)=2x+5g(x) = 2x + 5, find the value of xx where f(x)=g(x)f(x) = g(x).

Explanation**:**

Here, we're asked to find out where two different functions give the same result for the same value of xx. This is like finding the spot where two lines cross each other on a graph. To do this, we set the two functions equal to each other and solve for xx.


Solutions:


Problem 1: Evaluating a Function

infoNote

Question: Given the function f(x)=2x23x+4f(x) = 2x^2 - 3x + 4, find:

  1. f(1)f(1)
  2. f(2)f(-2)

For f(1)f(1):

  • First, replace xx with 11 in the function. This is like plugging 11 into the machine to see what comes out.

  • Plug in x=1x = 1: f(1)=2(1)23(1)+4f(1) = 2(1)^2 - 3(1) + 4

  • Now, do the math: 2(1)2=2,3(1)=3,23+4=32(1)^2 = 2, \quad -3(1) = -3, \quad 2 - 3 + 4 = 3

  • Answer: f(1)=3f(1) = 3. For f(2)f(-2):

  • Next, replace xx with 2-2 and do the same thing.

  • Plug in x=2x = -2: f(2)=2(2)23(2)+4f(-2) = 2(-2)^2 - 3(-2) + 4

  • Now, do the math: 2(2)2=8,3(2)=6,8+6+4=182(-2)^2 = 8, \quad -3(-2) = 6, \quad 8 + 6 + 4 = 18

  • Answer: f(2)=18f(-2) = 18. Explanation:

We replaced x x with the given numbers and followed the function's steps to see what comes out. This helps us understand how the function behaves for different inputs.


Problem 2: Finding Inputs for a Function

infoNote

Question: If f(x)=5x8f(x) = 5x - 8, find the value of xx when f(x)=17. f(x) = 17 .

To find xx when f(x)=17f(x) = 17:

  • Start by setting the function equal to 1717 because that's the result we're given. We need to figure out what number xx makes this happen.
  • Set up the equation: 5x8=175x - 8 = 17
  • First, add 88 to both sides to undo the subtraction. This gets us closer to isolating xx: 5x=255x = 25
  • Next, divide by 55 to find xx. We divide because xx is multiplied by 55, and dividing cancels this out: x=5x = 5
  • Answer: x=5x = 5. Explanation:

We worked backward from the result (17)(17) to find the starting number xx. By undoing each operation step by step, we were able to solve for xx.


Problem 3: Graphing a Linear Function

infoNote

Question: Graph the linear function y=2x+3y = -2x + 3 for xx values from 2-2 to 22.

Points to calculate:

  • We start by choosing different values for xx, and then we calculate the corresponding yy values. This helps us see where the function is on the graph.

  • Calculate for each value of xx:

    • When x=2x = -2 : y=2(2)+3=7y = -2(-2) + 3 = 7
    • When x=1x = -1 : y=2(1)+3=5y = -2(-1) + 3 = 5
    • When x=0x = 0 : y=2(0)+3=3y = -2(0) + 3 = 3
    • When x=1x = 1 : y=2(1)+3=1y = -2(1) + 3 = 1
    • When x=2x = 2 : y=2(2)+3=1y = -2(2) + 3 = -1
  • Next, plot these points on a graph. For example, plot the point (2,7)(-2, 7) by going to 2-2 on the x-axis and up to 77 on the y-axis.

    image
  • Finally, connect the dots with a straight line to complete the graph of the function. Explanation:

By calculating the yy values for different xx values, we can plot these points and connect them to draw the function. Since it's a linear function, the points will always line up in a straight line.


Problem 4: Finding the Roots of a Quadratic Function

infoNote

Question: Find the roots of the quadratic function f(x)=x25x+6f(x) = x^2 - 5x + 6.

  • The roots are the points where the function equals zero. So, we need to find out where f(x)=0f(x) = 0.
  • Set the equation: f(x)=x25x+6=0f(x) = x^2 - 5x + 6 = 0
  • To find the roots, we factorise the quadratic equation, which is like breaking it down into two simpler pieces.
  • Factorise: (x2)(x3)=0(x - 2)(x - 3) = 0
  • Now, solve for xx by setting each factor equal to zero:
    • x=2x = 2
    • x=3x = 3
  • Answer: The roots are x=2x = 2 and x=3x = 3. Explanation:

We found the roots by solving the equation where the function equals zero, which shows us where the graph crosses the x-axis. If factorizing is difficult, you can also use the quadratic formula to find the roots.


Problem 5: Solving a System of Linear Functions

infoNote

Question: If f(x)=3x+2f(x) = 3x + 2 and g(x)=2x+5g(x) = 2x + 5, find the value of xx where f(x)=g(x)f(x) = g(x).

  • We need to find the point where both functions give the same result. This is like finding where two lines cross on a graph.
  • Start by setting the functions equal to each other: 3x+2=2x+53x + 2 = 2x + 5
  • Next, move the xx terms to one side by subtracting 2x2x from both sides: 3x2x+2=53x - 2x + 2 = 5 x+2=5x + 2 = 5
  • Finally, subtract 22 from both sides to isolate xx: x=3x = 3
  • Answer: x=3x = 3. Explanation:

We set the two functions equal to find where they give the same result. By solving for xx, we find the value where the two lines intersect. If you input 33 into either function, they will both give the same output.


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