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Question Video: Finding the Variation Function of a Quadratic Function Mathematics

Determine the variation function 𝑉(β„Ž) for 𝑓(π‘₯) = βˆ’4π‘₯Β² βˆ’ 9π‘₯ + 9 at π‘₯ = βˆ’1.

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Video Transcript

Determine the variation function 𝑉 of β„Ž for 𝑓 of π‘₯ equals negative four π‘₯ squared minus nine π‘₯ plus nine at π‘₯ equals negative one.

In this question, we are given a quadratic function 𝑓 of π‘₯ and asked to find its variation function at a given value of π‘₯. To do this, we can start by recalling that the variation function 𝑉 of β„Ž of 𝑓 of π‘₯ at π‘₯ equals π‘Ž is given by 𝑉 of β„Ž equals 𝑓 of π‘Ž plus β„Ž minus 𝑓 of π‘Ž. In other words, we evaluate the function at π‘Ž and then subtract this from the function evaluated at π‘Ž plus β„Ž. The variation function 𝑉 of β„Ž tells us how much the function 𝑓 varies as the value of π‘₯ changes from π‘Ž to π‘Ž plus β„Ž. For example, if 𝑉 of β„Ž is positive, then the function increases its output as the input changes from π‘Ž to π‘Ž plus β„Ž. And if 𝑉 of β„Ž is negative, then it decreases its output.

We can use this definition to determine the variation function of the given quadratic at π‘₯ equals negative one. We have that 𝑓 of π‘₯ is negative four π‘₯ squared minus nine π‘₯ plus nine and π‘Ž equals negative one. We can substitute π‘Ž equals negative one into the formula for the variation function to obtain 𝑉 of β„Ž equals 𝑓 of negative one plus β„Ž minus 𝑓 of negative one. We can evaluate this in steps. Let’s start with 𝑓 of negative one plus β„Ž. We substitute this into the given function 𝑓 of π‘₯ to get negative four times negative one plus β„Ž all squared minus nine times negative one plus β„Ž plus nine.

We could follow the same process for 𝑓 evaluated at negative one; however, we can just directly evaluate this for simplicity. We substitute π‘₯ equals negative one into the function 𝑓 to get negative four times negative one squared minus nine times negative one plus nine. We can evaluate this to get 14. We can now substitute this into our expression for 𝑉 of β„Ž. We want to simplify this expression for the variation function. We can do this in stages. First, we can expand the exponent. We get one minus two β„Ž plus β„Ž squared. We need to multiply this by negative four.

Next, we can distribute negative nine over the parentheses. This gives us nine minus nine β„Ž. We can also evaluate the final two terms. We have nine minus 14 is equal to negative five. This gives us the following expression for the variation function. We can simplify further by distributing the factor of negative four over the parentheses. This gives us negative four plus eight β„Ž minus four β„Ž squared plus nine minus nine β„Ž minus five.

We now want to combine the like terms. We only have a single β„Ž squared term of negative four β„Ž squared. We can combine the β„Ž terms to obtain negative β„Ž. Finally, we have negative four plus nine minus five equals zero. This then gives us the variation function 𝑉 of β„Ž of 𝑓 at π‘₯ equals negative one is negative four β„Ž squared minus β„Ž.

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