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Question Video: Finding the Variation Function of a Quadratic Function Then Determining the Values of Its Constants Mathematics

Determine the variation function 𝑣(β„Ž) for 𝑓(π‘₯) = π‘Žπ‘₯Β² + 𝑏π‘₯ + 2 at π‘₯ = 1, and, from 𝑣(1/2) = 7/2 and 𝑓(1) = 6, determine the constants π‘Ž and 𝑏.

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Video Transcript

Determine the variation function 𝑣 of β„Ž for the function 𝑓 of π‘₯ is equal to π‘Žπ‘₯ squared plus 𝑏π‘₯ plus two at π‘₯ is equal to one, and from 𝑣 of one-half is equal to seven over two and 𝑓 of one is equal to six, determine the constants π‘Ž and 𝑏.

In this question, we’re asked to determine the variation function 𝑣 of β„Ž for an unknown quadratic function 𝑓 of π‘₯ at the value of π‘₯ is equal to one. This is the first part of the question. So, let’s start by doing this. And to do this, we’re going to need to recall what we mean by a variation function of a function 𝑓 of π‘₯ at a specific value. We can recall that the variation function 𝑣 of β„Ž for a function 𝑓 of π‘₯ at π‘₯ is equal to 𝑑 is given by 𝑓 evaluated at 𝑑 plus β„Ž minus 𝑓 evaluated at 𝑑. It’s a measure of how much the function changes when π‘₯ changes from 𝑑 to 𝑑 plus β„Ž.

And in our case, our function 𝑓 of π‘₯ is the quadratic π‘Žπ‘₯ squared plus 𝑏π‘₯ plus two, and our value of 𝑑 is equal to one. First, to determine the variation function 𝑣 of β„Ž, let’s start by substituting 𝑑 is equal to one. We get that 𝑣 of β„Ž is equal to 𝑓 evaluated at one plus β„Ž minus 𝑓 evaluated at one. So, to find an expression for our variation function, we’re going to need to substitute π‘₯ is equal to one plus β„Ž and π‘₯ is equal to one into the given function 𝑓 of π‘₯.

First, we substitute π‘₯ is equal to one plus β„Ž into the function 𝑓 of π‘₯. We get π‘Ž multiplied by one plus β„Ž squared plus 𝑏 times one plus β„Ž plus two. We then need to subtract 𝑓 evaluated at one. So, we need to subtract π‘Ž times one squared plus 𝑏 times one plus two. This gives us the following expression for our variation function. And we now need to simplify this expression. First, let’s start by distributing the exponent over the parentheses in our first term. We can do this by using the binomial formula or by using the FOIL method. In either case, we distribute this to get one plus two β„Ž plus β„Ž squared, and we need to multiply this by π‘Ž.

In our second term, we can distribute the value of 𝑏 over the parentheses. We multiply each term by 𝑏. 𝑏 times one is 𝑏 and 𝑏 times β„Ž is π‘β„Ž. We then add two to this value, and we can simplify further. We notice π‘Ž times one squared is equal to π‘Ž and 𝑏 times one is equal to 𝑏. Finally, we can distribute the negative over these parentheses. So, we subtract π‘Ž, we subtract 𝑏, and we subtract two, giving us the following expression for 𝑣 of β„Ž. And we can simplify this expression further. We can notice that we have 𝑏 minus 𝑏, which is equal to zero, and we also have two minus two, which is also equal to zero.

Let’s now distribute π‘Ž over the parentheses in our first term. We multiply each term inside the parentheses by π‘Ž. We get π‘Ž plus two π‘Žβ„Ž plus β„Ž squared, and we need to add π‘β„Ž and subtract π‘Ž. And we can continue simplifying. We have π‘Ž minus π‘Ž, which is equal to zero. And finally, we can notice the first and third term in this expression share a factor of β„Ž. So, taking this factor out then gives us that 𝑣 of β„Ž is equal to π‘Žβ„Ž squared plus β„Ž multiplied by two π‘Ž plus 𝑏.

Let’s now clear some space and move on to the second part of this question. We need to determine the values of the constants π‘Ž and 𝑏 by using two pieces of information. 𝑣 evaluated at one-half is equal to seven over two and 𝑓 evaluated at one is equal to six. So, let’s start with the first equation we’re given. 𝑣 evaluated at one-half is equal to seven over two. And we can find an expression for 𝑣 evaluated at one-half by substituting β„Ž is equal to one-half into our equation for 𝑣 of β„Ž.

Substituting β„Ž is equal to one-half into our function 𝑣 of β„Ž gives us that 𝑣 of one-half is equal to π‘Ž times one-half squared plus one-half multiplied by two π‘Ž plus 𝑏. And now, we can simplify. First, one-half squared is equal to one-quarter, so the first term is π‘Ž over four. And in the second term, we can distribute one-half over the parentheses to get π‘Ž plus 𝑏 over two. So, we have π‘Ž over four plus π‘Ž plus 𝑏 over two. And the first two terms π‘Ž over four plus π‘Ž simplifies to give us five π‘Ž over four. And remember, this needs to be equal to seven over two.

So, we have one equation involving π‘Ž and 𝑏. Let’s now clear some space and find the second equation by using the fact that 𝑓 evaluated at one is equal to six. We have that six is equal to 𝑓 evaluated at one. And we can find an expression for 𝑓 evaluated at one by substituting π‘₯ is equal to one into 𝑓 of π‘₯. This gives us that six is equal to π‘Ž times one squared plus 𝑏 multiplied by one plus two. And we can evaluate this; it simplifies to give us π‘Ž plus 𝑏 plus two. And remember, this needs to be equal to six. However, we can subtract two from both sides of the equation. And since six minus two is equal to four, we get that four is equal to π‘Ž plus 𝑏.

We now have a pair of simultaneous equations in terms of π‘Ž and 𝑏. And we can solve this in a few different ways. Let’s start by clearing some space. One way we can solve these simultaneous equations is by noticing in the second equation, the coefficients of π‘Ž and 𝑏 are both one. So, we can easily rearrange this equation to find an expression for 𝑏 in terms of π‘Ž or π‘Ž in terms of 𝑏. For example, if we subtract 𝑏 from both sides of the equation, we see that π‘Ž must be equal to four minus 𝑏. We can then substitute π‘Ž is equal to four minus 𝑏 into our first simultaneous equation. Doing this then gives us that seven over two is equal to five times four minus 𝑏 over four plus 𝑏 over two.

We now have an equation entirely in terms of 𝑏. So, we can solve for 𝑏 by isolating 𝑏 on one side of the equation. To do this, let’s start by distributing five over the parentheses. We multiply each term in the parentheses by five. We get 20 minus five 𝑏, and we need to divide all of this by four. So, we have seven over two is equal to 20 minus five 𝑏 all over four plus 𝑏 over two. We can simplify our equation by multiplying both sides of the equation by the lowest common multiple of the denominators. In this case, that’s four. So, we multiply both sides of the equation by four to remove the denominators. Multiplying both sides of the equation by four gives us that 14 is equal to 20 minus five 𝑏 plus two 𝑏.

And now we can just solve this equation for 𝑏. Negative five 𝑏 plus two 𝑏 is negative three 𝑏. We can rearrange the equation to get that three 𝑏 is equal to six and then divide both sides of the equation through by three. We see that 𝑏 is equal to two. And we can verify this by substituting 𝑏 is equal to two back into our equation only in terms of 𝑏. We can also use this value of 𝑏 to determine the value of π‘Ž since remember, π‘Ž is equal to four minus 𝑏.

So, substituting 𝑏 is equal to two into this equation gives us that π‘Ž is equal to four minus two, which simplifies to give us two. And now that we’ve shown that π‘Ž is equal to two and 𝑏 is equal to two, we could substitute the values of these constants into our function 𝑣 of β„Ž. However, it’s not necessary. We can just say that π‘Ž is two and 𝑏 is two.

Therefore, we were able to show the variation function 𝑣 of β„Ž for 𝑓 of π‘₯ is equal to π‘Žπ‘₯ squared plus 𝑏π‘₯ plus two at π‘₯ is equal to one is 𝑣 of β„Ž is equal to π‘Žβ„Ž squared plus β„Ž times two π‘Ž plus 𝑏. And if 𝑣 of one-half is equal to seven over two and 𝑓 of one is equal to six, then π‘Ž is equal to two and 𝑏 is equal to two.

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