Question Video: Finding the Unknown Coefficient given a Linear Function and a Point on It | Nagwa Question Video: Finding the Unknown Coefficient given a Linear Function and a Point on It | Nagwa

Question Video: Finding the Unknown Coefficient given a Linear Function and a Point on It Mathematics • Third Year of Preparatory School

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If (βˆ’10, βˆ’5) belongs to the set of the function 𝑓, where 𝑓(π‘₯) = π‘šπ‘₯ + 15, find the value of π‘š.

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Video Transcript

If negative 10, negative five belongs to the set of the function 𝑓, where 𝑓 of π‘₯ equals π‘šπ‘₯ plus 15, find the value of π‘š.

Let’s begin by recalling what it means for an ordered pair to belong to the set of a function 𝑓. If an ordered pair belongs to the set of the function 𝑓, then we can represent this ordered pair as π‘₯, 𝑓 of π‘₯. π‘₯ is essentially the values that we input to that function and the set of all π‘₯-values is said to be the domain of that function. Then, for any values of π‘₯ that we substitute in, 𝑓 of π‘₯ is the value we get out; it’s the output. The set of all 𝑓 of π‘₯ values then is said to be the range of the function.

With this in mind, if we know that the ordered pair negative 10, negative five belongs to the set of the function, we know that when π‘₯ equals negative 10, 𝑓 of π‘₯, the output, is actually 𝑓 of negative 10, which is equal to negative five. So, substituting π‘₯ equals negative 10 into our function, we get 𝑓 of negative 10 is π‘š times negative 10 plus 15. And that simplifies to negative 10π‘š plus 15. But remember, we said 𝑓 of negative 10 is negative five. So, we can form and solve an equation for π‘š; that is, negative five equals negative 10π‘š plus 15.

We’ll then solve for π‘š by subtracting 15 from both sides of our equation. Negative five minus 15 is negative 20. So, we get negative 20 equals negative 10π‘š. Finally, we’re going to divide through by negative 10. Negative 20 divided by negative 10 is positive two. And so, we find that π‘š is equal to two.

Now, it’s worth noting that this means our function 𝑓 of π‘₯ can be alternatively written as two π‘₯ plus 15. We could then check our answer by resubstituting our values back into the function and checking we get the expected output, in other words, by evaluating 𝑓 of negative 10. Well, that’s two times negative 10 plus 15, which is negative 20 plus 15 or negative five as we expected. So, the value of π‘š in this occasion is two.

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