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5 posts
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state the turning point and sketch the graph of the following.
y = x^2 + 2
For this, you can't do the quadratic graph method like
when y =0. x =___
when x=0, y= ____
find turning point
find endpoints
when i sub y = 0. x^2 + 2 = 0
shift 2 to right-hand side and you get
x^2 = -2
i got stuck here cuz you cant square root a negative value. so what other method can i use?
the next question :
state the line of symmetry and sketch the graph of the following
y = -x^2 - 3. how do you get the line of symmetry? by looking at it, i can tell that it's 0 but is there anyone who can explain why?
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again, the graph x^2 + 2 does not intercept the x axis.
when x = 0, y =2. so you can expect that it is above the x-axis, but also intercepting the y-axis.
to find turning point, you can use dy/dx. at turning point, dy/dx = o. thus, find the value of x and sub it in the eqn y = x^2 +2 to find the corresponding y-value. if you have not learn differentiation, you can use the visual method to approach the question.
For example, y = x^2 +2 implies that a x^2 graph (centre at the 0,0 coordinate) moved 2 units up along the y-axis. so the turning point would be from (0,0) to (0,2).
for the next question, -x^2 graph implies that the graph is a x^2 graph, but reflected across the x-axis. i.e it is a x^2 graph that looks like you drawn it upside down and below the x-axis.
as the above question, the reflected x^2 graph should also cut the origin at (0,2). since the eqn given is y= -x^2 -3, this implies that the whole graph shifted 3 units down along the y-axis. thus the turning point of the graph is (0, -3).
knowing the property of the graph, i.e the shape of the graph, you should be able to deduce that the line of symmetry is the x-coordinate of the turning point, i.e the line of symmetry is x=0.
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Originally posted by starmoonsun:
again, the graph x^2 + 2 does not intercept the x axis.
when x = 0, y =2. so you can expect that it is above the x-axis, but also intercepting the y-axis.
to find turning point, you can use dy/dx. at turning point, dy/dx = o. thus, find the value of x and sub it in the eqn y = x^2 +2 to find the corresponding y-value. if you have not learn differentiation, you can use the visual method to approach the question.
For example, y = x^2 +2 implies that a x^2 graph (centre at the 0,0 coordinate) moved 2 units up along the y-axis. so the turning point would be from (0,0) to (0,2).
for the next question, -x^2 graph implies that the graph is a x^2 graph, but reflected across the x-axis. i.e it is a x^2 graph that looks like you drawn it upside down and below the x-axis.
as the above question, the reflected x^2 graph should also cut the origin at (0,2). since the eqn given is y= -x^2 -3, this implies that the whole graph shifted 3 units down along the y-axis. thus the turning point of the graph is (0, -3).
knowing the property of the graph, i.e the shape of the graph, you should be able to deduce that the line of symmetry is the x-coordinate of the turning point, i.e the line of symmetry is x=0.
How do you know the graph is centred at (0,0)?
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star means that for a y=x^2 graph, the minimum point/turning point is at the (0,0) coordinate
but when you plot y=x^2 + 2, the graph moves up by 2 units, thus the turning point moves up by 2 units to (0,2)
How you know it is at (0,0) => You have to memorize and know the different basic graphs
Regards,
Eagle
Owner of Strategic TuitionEdited by eagle 09 Aug `08, 4:28PM
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