Video: Determining the Variation Function of a Quadratic Function and Finding the Value of One of Its Unknowns

Determine the variation function 𝑉(β„Ž) for 𝑓(π‘₯) = βˆ’π‘₯Β² + π‘Žπ‘₯ + 17 at π‘₯ = βˆ’1. Additionally, find π‘Ž if 𝑉(4/9) = 11/6.

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Video Transcript

Determine the variation function 𝑉 of β„Ž for 𝑓 of π‘₯ is equal to negative π‘₯ squared plus π‘Žπ‘₯ plus 17 at π‘₯ is equal to negative one. Additionally, find π‘Ž if 𝑉 of four over nine is 11 over six.

We’re given the function 𝑓 of π‘₯ is negative π‘₯ squared plus π‘Žπ‘₯ plus 17 and asked to find the variation function for this at π‘₯ is negative one. To do this, we recall that for a function 𝑓 of π‘₯ at π‘₯ is equal to 𝛼, the variation function 𝑉 of β„Ž is 𝑓 of 𝛼 plus β„Ž minus 𝑓 of 𝛼, where β„Ž is the change in π‘₯ from π‘₯ is equal to 𝛼. Once we found our function 𝑉 of β„Ž, we’ll then substitute β„Ž is four over nine to find the value of π‘Ž in our function 𝑓. So we’re given 𝑓 of π‘₯ is negative π‘₯ squared plus π‘Žπ‘₯ plus 17. With π‘₯ is negative one, this is equal to our 𝛼. And substituting 𝛼 is negative one, we have 𝑉 of β„Ž is 𝑓 of negative one plus β„Ž minus 𝑓 of negative one.

And now evaluating our function 𝑓 at π‘₯ is equal to negative one plus β„Ž, we have negative negative one plus β„Ž squared plus π‘Ž times negative one plus β„Ž plus 17. That is, 𝑓 of negative one plus β„Ž is equal to negative one minus two β„Ž plus β„Ž squared minus π‘Ž plus π‘Žβ„Ž plus 17. And multiplying our parentheses by negative one and collecting like terms, we have negative β„Ž squared plus β„Ž times π‘Ž plus two plus 16 minus π‘Ž. Now, evaluating 𝑓 at π‘₯ is equal to negative one, we have negative negative one squared plus π‘Ž times negative one plus 17, that is, negative one minus π‘Ž plus 17, which is 16 minus π‘Ž. And substituting our two results into our function 𝑉 of β„Ž, we see that 16 minus π‘Ž minus 16 minus π‘Ž is equal to zero so that 𝑉 of β„Ž is equal to negative β„Ž squared plus β„Ž times π‘Ž plus two.

Our variation function for 𝑓 of π‘₯ is equal to negative π‘₯ squared plus π‘Žπ‘₯ plus 17 at π‘₯ is negative one is equal to 𝑉 of β„Ž is negative β„Ž squared plus β„Ž times π‘Ž plus two. Now we’re given that 𝑉 of four over nine is equal to 11 over six. And this means that if we substitute β„Ž is equal to four over nine into 𝑉 of β„Ž, this should equal 11 over six. And this means that negative four over nine squared plus four over nine times π‘Ž plus two is 11 over six. And we’re going to use this to find the value of π‘Ž. Evaluating this gives us negative 16 over 81 plus four π‘Ž over nine plus eight over nine is equal to 11 over six. And now if we add 16 over 81 and subtract eight over nine from both sides and multiplying both sides by nine over four, we can isolate π‘Ž on the left-hand side.

Canceling through our parentheses, we then have π‘Ž is equal to 33 over eight plus four over nine minus two, which evaluates to 2.5694 and so on. So with our variation function 𝑉 of β„Ž is negative β„Ž squared plus β„Ž times π‘Ž plus two, we have a value of π‘Ž equal to 2.57 to two decimal places.

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