Solve the equation
$x^2 - x - 6 = 0.$
The graphs of four quadratic functions are shown below - Leaving Cert Mathematics - Question (a)(b)(c) - 2014
Question (a)(b)(c)
Solve the equation
$x^2 - x - 6 = 0.$
The graphs of four quadratic functions are shown below.
Graph A Graph B
Graph C Graph D
Which of the graphs above is t... show full transcript
Worked Solution & Example Answer:Solve the equation
$x^2 - x - 6 = 0.$
The graphs of four quadratic functions are shown below - Leaving Cert Mathematics - Question (a)(b)(c) - 2014
Step 1
Solve the equation $x^2 - x - 6 = 0$
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Answer
To solve the equation x2−x−6=0, we can factor it:
(x+2)(x−3)=0
This gives us two solutions:
x+2=0⟹x=−2
x−3=0⟹x=3
Therefore, the solutions are x=−2 and x=3.
Step 2
Which of the graphs above is that of the function $f: x \to x^2 - x - 6$?
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Answer
The correct graph for the function f(x)=x2−x−6 can be identified by finding the vertex and the intercepts. The vertex lies at x=2a−b=21. The function opens upwards, and the curve intersects the x-axis at x=−2 and x=3. Upon examining the provided graphs,
Graph D
accurately depicts these features.
Step 3
(i) $h(x) = g(x) + 2$
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Answer
To sketch h(x), we take the graph of g(x) and shift it up by 2 units. Therefore, the entire graph of g(x) will be displaced vertically upwards. This includes moving the vertex from (1, 1) to (1, 3). The new points after the upward shift will represent the graph of h(x).
Step 4
(ii) $k(x) = g(x + 2)$
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Answer
To sketch k(x), we need to shift the graph of g(x) horizontally to the left by 2 units. This means if the original vertex of g(x) is at (1, 1), the vertex of k(x) will now be at (-1, 1). Plot the new vertex and ensure that the shape of the parabola remains consistent while adjusting the x-coordinates accordingly.
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