Solving for Function with Given Indicated Values
I'm relatively new to the world of functions and solving them with given values, so I'm completely lost here. The question is as follows: "Given the following function, find the indicated values: $f(x)$ = $x^2$ - $5x$ + $1$
How would I solve this problem? Where would I even begin?
$\endgroup$2 Answers
$\begingroup$Presumably you are given a list of $x$ values. You are supposed to fill in a table, evaluating $f(x)$ at each of those values. Something like $$\begin {array}{c| c} x&f(x) \\\hline 0&1\\1&-3\\2&-5 \end {array}$$
$\endgroup$ 3 $\begingroup$I don't know what you mean by "solving" the function. I think they gave you a function: $$f(x)=x^2-5x+1$$ and told you to find $f(7)$ or something. In that case, I will try to explain not only how to find $f(7)$, but how functions are in general (since you said you were new to functions).
First, you find $f(7)$ by plugging in $7$ to every $x$. That means that $f(7)=7^2-5(7)+1$. Then you evaluate it. $f(7)=49-35+1=15$. That means $f(7)=15$. I will talk about how this is important later.
A function is like a machine. $f$ is the name of the machine (it can be $g$ or $h$ or whatever you like), and $x$ is the thing you input into the machine. Our machine $f(x)=x^2-5x+1$ is saying this: "Throw some $x$ into me, and I will give you $x^2-5x+1$!" When you throw a $7$ into the machine, the machine gives you $7^2-5(7)+1$, which is $15$.
Our machine can only output one number. If it has to choose between $2$ numbers, it will not know what to do. For example, if the machine handles the equation $x^2+y^2=25$, and you throw an $x$ value of $4$ into the equation and ask it to solve for $y$, it will think:
"Okay, $4^2+y^2=25$. So $y^2=25-16=9$. $y=\pm 3$. Which one?! $3$ or $-3$?!!" $\text{*self destructs*}$
Every function must have only one y-value for every x-value. Our equation $x^2+y^2=25$ is not a function! because for an x-value of $4$, there were two possible y-values; $3$ and $-3$. If you are curious, the graph of the equation above is of a circle having a centre $(0,0)$ and radius $5$.
Every point on the graph of a function must be in the form $(x, f(x))$. Let's take our first example $f(x)=x^2-5x+1$. We found out that $f(7)=15$. A point on the graph of $f(x)=x^2-5x+1$ is going to be $(7, f(7))$, which is $(7,15)$. On the other hand, $(7,14)$ is not on the graph! because $f(7)\neq 14$.
You will learn a lot more about functions and their graphs later on, but I think that, for now, this is all you need to know.
OP has specified that he was given nothing else except $f(x)=x^2-5x+1$. He has to "solve for the function", so I am going to guess that he wants to find the roots of the function.
A root is/are the point(s) where the graph of the function intersects the x-axis. In other words, $f(x)=0$. This is also referred to as a zero, or an x-intercept. If you think about it, this is a system of equations with two equations and two unknowns. $$\begin{cases} f(x)=x^2-5x+1 \\ f(x)=0 \\ \end{cases}$$ We know $f(x)=0$. Therefore $0=x^2-5x+1$. This is called solving by substitution. We are making the substitution $f(x)=0$. How do we solve $x^2-5x+1=0$? There is a very well known formula called the Quadratic Formula. It states:
For any equation in the form $ax^2+bx+c=0$, where $a$, $b$ and $c$ are constants and $a\neq 0$, $x$ is equal to: $$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$$
We can apply this formula to our equation $x^2-5x+1=0$. We know that $a=1$, $b=-5$, and $c=1$. Therefore: $$x=\frac{-(-5)\pm\sqrt{(-5)^2-4(1)(1)}}{2(1)}$$ $$x=\frac{5\pm\sqrt{25-4}}{2}$$ $$x=\frac{5\pm\sqrt{21}}{2}$$ So the solutions are: $$x=\frac{5+\sqrt{21}}{2}$$ $$x=\frac{5-\sqrt{21}}{2}$$ Now we have our roots. We say that, "The roots are $\dfrac{5+\sqrt{21}}{2}$ and $\dfrac{5-\sqrt{21}}{2}$."
Perhaps you also want to find the y-intercept. This is much easier than finding the roots. The y-intercept is where the graph intersects with the y-axis. In other words, the y-intercept is where the x-coordinate is $0$. Therefore the y-intercept will have the coordinates $(0, y)$. Remember what I said about every point on a function having coordinates $(x, f(x))$? We know that $x=0$, so the y-intercept is going to be $(0, f(0))$. We just need to find $f(0)$, which is easy. $$f(0)=0^2-5(0)+1$$ $$f(0)=1$$ Therefore the y-intercept has coordinates $(0,1)$. We just say that, "The y-intercept is $1$."
If you want to know how to derive the quadratic formula, here is how I do it. We have our standard equation: $$ax^2+bx+c=0$$ Multiply by $4a$ on both sides: $$4a^2x^2+4abx+4ac=0$$ Subtract $4ac$ from both sides: $$4a^2x^2+4abx=-4ac$$ There is a formula that $(a+b)^2=a^2+2ab+b^2$. We can then the left hand side into a expression of the form $(a+b)^2$ by adding $b^2$ to both sides. $$4a^2x^2+4abx+b^2=b^2-4ac$$ $$(2ax)^2+2(2ax\cdot b)+(b)^2=b^2-4ac$$ $$(2ax+b)^2=b^2-4ac$$ Taking the square root of both sides: $$2ax+b=\pm\sqrt{b^2-4ac}$$ $$2ax=-b\pm\sqrt{b^2-4ac}$$ $$\displaystyle \boxed{x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}}$$ And that is how you derive the quadratic formula.
I could go on and on about parabolas (the graph of any quadratic equation is a parabola; a U shape), talking about foci, directrix, etc. But I feel like you are getting bored! So I am going to stop here, because this is basically functions in a nutshell.
Hope I helped! $\endgroup$ 2
