#Functions help pls :)

31 messages · Page 1 of 1 (latest)

hardy sand
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I’m not sure how to approach this questions

topaz knot
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Which part

dark thicket
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You know that a quadratic has a turning point

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If its positive then all values of the quadratic are greater than the value at its turning point

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So the range will be y≥k where k is the y-oordinate of the quadratic

hardy sand
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Complete the square got me (x+2a)^2+ a^2- 4

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Or is that incorrect

dark thicket
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It's wrong

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(2a)²=4a² right

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So you subtract 4a² at the end

hardy sand
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Ur right

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Thanks

dark thicket
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For part ii:
Evaluate g(3) and sub this expression into the xs of f(x)
You should get a quadratic in a
Solve to find for a
Now you have a linear function, which you can just take the inverse function of and equate to x, which is then solving a linear equation

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Note that you might get 2 solutions to a so you might have 2 values of x

hardy sand
dark thicket
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Ok so if you sketch your general qudratic

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And then plot the point (-2a, -3a²) on it

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That'll be on ur turning point

hardy sand
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Yh

dark thicket
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Range is all the y values that are function can take

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A positive quadratic can take any value above and equal to its turning point

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As you should be able to convince yourself by a sketch

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So you take the y values of the coordinates of the turning point which will be the lowest value that the function can taks

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And it can take every value greater than that

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So the range is y≥-3a²

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Or f(x)≥-3a²

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Both are valid

hardy sand
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Thanks man appreciate the help

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Peace✌🏽