Practice Problems (Junior Cert Mathematics): Revision Notes

Practice Problems

Problems:

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Problem 1: Evaluating a Function

Explanation**:**

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

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Problem 2: Finding Inputs for a Function

Explanation**:**

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

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Problem 3: Graphing a Linear Function

Explanation**:**

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

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Problem 4: Finding the Roots of a Quadratic Function

Explanation**:**

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

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Problem 5: Solving a System of Linear Functions

Explanation**:**

Here, we're asked to find out where two different functions give the same result for the same value of $x$. 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 $x$.

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Solutions:

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Problem 1: Evaluating a Function

For $f(1)$:

• First, replace $x$ with $1$ in the function. This is like plugging $1$ into the machine to see what comes out.

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

• Now, do the maths: $2(1)^2 = 2, \quad -3(1) = -3, \quad 2 - 3 + 4 = 3$

• Answer: $f(1) = 3$. For $f(-2)$:

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

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

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

• Answer: $f(-2) = 18$. Explanation:

We replaced $x$ 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.

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Problem 2: Finding Inputs for a Function

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

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

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

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Problem 3: Graphing a Linear Function

Points to calculate:

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

• Calculate for each value of $x$:

  • When $x = -2$ : $y = -2(-2) + 3 = 7$
  • When $x = -1$ : $y = -2(-1) + 3 = 5$
  • When $x = 0$ : $y = -2(0) + 3 = 3$
  • When $x = 1$ : $y = -2(1) + 3 = 1$
  • When $x = 2$ : $y = -2(2) + 3 = -1$

• Next, plot these points on a graph. For example, plot the point $(-2, 7)$ by going to $-2$ on the x-axis and up to $7$ on the y-axis.

  

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

By calculating the $y$ values for different $x$ 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.

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Problem 4: Finding the Roots of a Quadratic Function

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

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Problem 5: Solving a System of Linear Functions

• 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: $3x + 2 = 2x + 5$
• Next, move the $x$ terms to one side by subtracting $2x$ from both sides: $3x - 2x + 2 = 5$ $x + 2 = 5$
• Finally, subtract $2$ from both sides to isolate $x$: $x = 3$
• Answer: $x = 3$. Explanation:

We set the two functions equal to find where they give the same result. By solving for $x$, we find the value where the two lines intersect. If you input $3$ into either function, they will both give the same output.

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