Part of the graph of the function $y = x^2 + ax + b$, where $a, b \in \mathbb{Z}$, is shown below. The points $R(2, 3)$ and $S(-5, -4)$ are on the curve. (a) Use the given points to form two equations in $a$ and $b$. (b) Solve your equations to find the value of $a$ and the value of $b$. (c) Write down the co-ordinates of the point where the curve crosses the y-axis. (d) By solving an equation, find the points where the curve crosses the x-axis. Give each answer correct to one decimal place.

Junior Cycle Mathematics ALGEBRA - Factorising expressions: Part of the graph of the functio

Part of the graph of the function $y = x^2 + ax + b$, where $a, b \in \mathbb{Z}$, is shown below - Junior Cycle Mathematics - Question 10 - 2011

Question 10

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Part of the graph of the function $y = x^2 + ax + b$, where $a, b \in \mathbb{Z}$, is shown below. The points $R(2, 3)$ and $S(-5, -4)$ are on the curve. (a) Use t... show full transcript

Worked Solution & Example Answer:Part of the graph of the function $y = x^2 + ax + b$, where $a, b \in \mathbb{Z}$, is shown below - Junior Cycle Mathematics - Question 10 - 2011

Step 1

Use the given points to form two equations in $a$ and $b$.

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Answer

Substituting the point $R(2, 3)$ into the equation gives:

$3 = (2)^2 + 2a + b$
Simplifying this:
$3 = 4 + 2a + b$
Which leads to:
$2a + b = -1$

For the point $S(-5, -4)$ :

$-4 = (-5)^2 + a(-5) + b$
Simplifying this:
$-4 = 25 - 5a + b$
Which leads to:
$-5a + b = 29$

Step 2

Solve your equations to find the value of $a$ and the value of $b$.

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Answer

We have the equations:

$2a + b = -1$
$-5a + b = 29$

To eliminate $b$ , subtract the first equation from the second:

$(-5a + b) - (2a + b) = 29 - (-1)$
This simplifies to:

$-7a = 30$
Thus,
$a = -\frac{30}{7} = 4$ .

Substituting $a = 4$ back into one of the equations:

$2(4) + b = -1$
This gives: $8 + b = -1$
So,
$b = -9$ .

Step 3

Write down the co-ordinates of the point where the curve crosses the y-axis.

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Answer

The curve crosses the y-axis when $x = 0$ . Substituting $x = 0$ into the equation gives:

y = (0)^2 + 4(0) - 9 = -9.

Thus, the coordinates are $(0, -9)$ .

Step 4

By solving an equation, find the points where the curve crosses the x-axis.

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Answer

The curve crosses the x-axis when $y = 0$ , so we solve:

$0 = x^2 + 4x - 9$ .

Using the quadratic formula:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$
Setting $a = 1, b = 4, c = -9$ , we have:

$x = \frac{-4 \pm \sqrt{16 + 36}}{2(1)}$
This simplifies to: $x = \frac{-4 \pm \sqrt{52}}{2}$
$x = \frac{-4 \pm 2\sqrt{13}}{2}$
Thus, $x = -2 \pm \sqrt{13}$ .

Calculating the approximate values:

$x \approx -2 + 3.6 \approx 1.6$
$x \approx -2 - 3.6 \approx -5.6$ .

Thus, the points where the curve crosses the x-axis are approximately $1.6$ and $-5.6$ .

Part of the graph of the function $y = x^2 + ax + b$, where $a, b \in \mathbb{Z}$, is shown below - Junior Cycle Mathematics - Question 10 - 2011

Worked Solution & Example Answer:Part of the graph of the function $y = x^2 + ax + b$, where $a, b \in \mathbb{Z}$, is shown below - Junior Cycle Mathematics - Question 10 - 2011

Use the given points to form two equations in $a$ and $b$.

Solve your equations to find the value of $a$ and the value of $b$.

Write down the co-ordinates of the point where the curve crosses the y-axis.

By solving an equation, find the points where the curve crosses the x-axis.

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