Find The Cubic Polynomial Equation & Its Y-Intercept

Hey math whizzes and future mathematicians! Ever stared at a graph and wondered what magical equation conjures it up? Well, today, we're diving deep into the fascinating world of cubic polynomial functions. We've got a specific puzzle to solve: we need to find the equation of a cubic polynomial function given its zeroes and a point it passes through. Plus, we'll uncover its $y$-intercept. So grab your notebooks, guys, because this is going to be a fun ride!

Unpacking the Cubic Polynomial Function

Alright, let's get down to business. A cubic polynomial function is basically a function that includes a term with $x^3$ as its highest power. Think of it as a fancy, wiggly line on a graph that can change direction a couple of times. The zeroes of a function are super important – they're the $x$-values where the function's output ($y$) is zero. Graphically, these are the points where the graph crosses or touches the $x$-axis. If a cubic polynomial function has zeroes at $x=r_1$, $x=r_2$, and $x=r_3$, then we can express the function in its factored form like this: $f(x) = a(x - r_1)(x - r_2)(x - r_3)$. Here, '$a$' is a constant that tells us about the 'stretch' or 'compression' of the graph, and whether it opens upwards or downwards. Our mission, should we choose to accept it, is to find the specific values of $a$, $r_1$, $r_2$, and $r_3$ for the function described in our problem. We are given that the zeroes are $(2,0)$, $(3,0)$, and $(5,0)$. This directly tells us that $r_1 = 2$, $r_2 = 3$, and $r_3 = 5$. So, our function's general form now looks like $f(x) = a(x - 2)(x - 3)(x - 5)$. The only piece of the puzzle missing is the value of '$a$'. Don't worry, we've got another clue coming right up!

Using a Point to Find 'a'

Now, to nail down that elusive constant '$a$', we need to use the extra piece of information the problem gives us: the function passes through the coordinate $(0, -5)$. What does this mean, exactly? It means when $x=0$, the value of the function, $f(x)$, is $-5$. We can plug these values into our current equation: $f(x) = a(x - 2)(x - 3)(x - 5)$. Substituting $x=0$ and $f(x)=-5$, we get: $-5 = a(0 - 2)(0 - 3)(0 - 5)$. Let's crunch those numbers: $-5 = a(-2)(-3)(-5)$. Multiplying the numbers inside the parentheses, we get $(-2) imes (-3) = 6$, and then $6 imes (-5) = -30$. So, the equation becomes $-5 = a(-30)$. To find '$a$', we just need to divide both sides by $-30$: a = rac{-5}{-30}. Simplifying this fraction, we get a = rac{1}{6}. Awesome! We've found our '$a$'!

Constructing the Final Equation

With the value of '$a$' in hand, we can now write the complete equation for our cubic polynomial function. We substitute a = rac{1}{6} back into our factored form: f(x) = rac{1}{6}(x - 2)(x - 3)(x - 5). This is the equation of the cubic polynomial function in its factored form. Sometimes, you might be asked to expand this into the standard polynomial form, which looks like $f(x) = Ax^3 + Bx^2 + Cx + D$. To do that, you'd multiply out the factors $(x-2)(x-3)(x-5)$ first, and then multiply the result by rac{1}{6}. Let's do that expansion: First, multiply $(x-2)(x-3)$: $x^2 - 3x - 2x + 6 = x^2 - 5x + 6$. Now, multiply this by $(x-5)$: $(x^2 - 5x + 6)(x - 5) = x^2(x-5) - 5x(x-5) + 6(x-5) = x^3 - 5x^2 - 5x^2 + 25x + 6x - 30 = x^3 - 10x^2 + 31x - 30$. Finally, multiply the entire result by rac{1}{6}: f(x) = rac{1}{6}(x^3 - 10x^2 + 31x - 30) = rac{1}{6}x^3 - rac{10}{6}x^2 + rac{31}{6}x - rac{30}{6}. Simplifying the coefficients, we get f(x) = rac{1}{6}x^3 - rac{5}{3}x^2 + rac{31}{6}x - 5. So, the equation in standard form is f(x) = rac{1}{6}x^3 - rac{5}{3}x^2 + rac{31}{6}x - 5. Pretty neat, huh?

What's the Y-Intercept? Let's Find Out!

Now for the second part of our challenge: finding the $y$-intercept. What exactly is a $y$-intercept? It's the point where the graph of the function crosses the $y$-axis. And what's special about any point on the $y$-axis? Its $x$-coordinate is always zero! So, to find the $y$-intercept, we simply need to evaluate our function at $x=0$. We already actually did this when we were trying to find the value of '$a$'! Remember? We were given that the function passes through the coordinate $(0, -5)$. This is the $y$-intercept! The $y$-coordinate where the graph crosses the $y$-axis is $-5$. Alternatively, if you look at the standard form of the polynomial, $f(x) = Ax^3 + Bx^2 + Cx + D$, the constant term '$D$' is always the $y$-intercept. In our expanded equation, f(x) = rac{1}{6}x^3 - rac{5}{3}x^2 + rac{31}{6}x - 5, the constant term is $-5$. So, the $y$-intercept is indeed $-5$. This makes perfect sense because when you plug $x=0$ into the standard form, all the terms with $x$ become zero, leaving only the constant term $D$. So, $f(0) = A(0)^3 + B(0)^2 + C(0) + D = D$. Therefore, the $y$-intercept is always the constant term in the standard form of a polynomial. We found our constant term to be $-5$, confirming our earlier finding. It's always a good sign when different methods lead to the same answer, right guys?

Conclusion: Putting It All Together

So there you have it! We successfully determined the equation of the cubic polynomial function using its zeroes and a given point. The equation in factored form is f(x) = rac{1}{6}(x - 2)(x - 3)(x - 5), and in standard form, it's f(x) = rac{1}{6}x^3 - rac{5}{3}x^2 + rac{31}{6}x - 5. And the $y$-intercept? That's the point where the graph crosses the $y$-axis, which we found to be $-5$. This means the graph passes through the point $(0, -5)$. Looking at the multiple-choice options provided (A. -5, B. -2, C. 3, D. 5), the correct answer for the $y$-intercept is A. -5. Keep practicing these types of problems, and soon you'll be spotting cubic polynomials and their intercepts like a pro! Math is all about breaking down complex problems into smaller, manageable steps, and this problem was a perfect example of that. Keep exploring, keep questioning, and most importantly, keep having fun with mathematics!