Question Video: Finding the Value of the Variation Function of a Quadratic Function | Nagwa Question Video: Finding the Value of the Variation Function of a Quadratic Function | Nagwa

Question Video: Finding the Value of the Variation Function of a Quadratic Function Mathematics • Second Year of Secondary School

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If 𝑉 is the variation function for 𝑓(π‘₯) = π‘₯Β² βˆ’ 4π‘₯ + 2, what is 𝑉(βˆ’0.2) when π‘₯ = 8?

02:45

Video Transcript

If 𝑉 is the variation function for 𝑓 of π‘₯ is equal to π‘₯ squared minus four π‘₯ plus two, what is 𝑉 of negative 0.2 when π‘₯ is equal to eight?

We’re given a quadratic function 𝑓 of π‘₯ is π‘₯ squared minus four π‘₯ plus two and asked to find the value of the variation function if π‘₯ changes from π‘₯ is equal to eight by an amount of negative 0.2. The first thing we need to do then is to find the variation function 𝑉 of β„Ž. We can do this using the definition for a function 𝑓 of π‘₯ where its variation function at π‘₯ is equal to π‘Ž is given by 𝑉 of β„Ž is 𝑓 of π‘Ž plus β„Ž minus 𝑓 of π‘Ž. And that’s where β„Ž is the change in π‘₯. In our case, our function 𝑓 of π‘₯ is π‘₯ squared minus four π‘₯ plus two. Our given value of π‘₯ is eight so that π‘Ž is equal to eight so that our variation function 𝑉 of β„Ž is 𝑓 of eight plus β„Ž minus 𝑓 of eight.

There are two ways we could approach this problem. We know that our change in π‘₯ is negative 0.2, and this means that β„Ž is negative 0.2. And we could substitute this directly into our equation for 𝑉 of β„Ž. Alternatively, we could first find 𝑉 of β„Ž, the variation function in terms of β„Ž, and then substitute β„Ž is negative 0.2. We’re going to use the second method to find the variation function 𝑉 of β„Ž. And to do this, we’re first going to substitute π‘₯ is equal to eight plus β„Ž into our function 𝑓 of π‘₯. And this gives us eight plus β„Ž squared minus four times π‘Ž plus β„Ž plus two. Distributing our parentheses, this gives us 64 plus 16β„Ž plus β„Ž squared minus 32 minus four β„Ž plus two. And collecting like terms, this gives us eight squared plus 12β„Ž plus 34.

Now, to find 𝑓 of eight, we substitute π‘₯ equal to eight into our function 𝑓 of π‘₯, which gives us eight squared minus four times eight plus two. And this evaluates to 34. Now, with these two results into our function 𝑉 of β„Ž, we have eight squared plus 12β„Ž plus 34 minus 34, that is, β„Ž squared plus 12β„Ž. And this is our variation function 𝑉 of β„Ž for 𝑓 of π‘₯ at π‘₯ is equal to eight. Now, to find 𝑉 of negative 0.2, we substitute negative 0.2 in place of β„Ž. This gives us negative 0.2 squared plus 12 times negative 0.2. That is 0.04 minus 2.4, which is negative 2.36. 𝑉 of negative 0.2 for 𝑓 of π‘₯ is equal to π‘₯ squared minus four π‘₯ plus two when π‘₯ is equal to eight is therefore negative 2.36.

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