Find The Cubic Polynomial: Zeros & A Point Given

Hey guys, ever found yourself staring at a math problem that feels like a puzzle? Well, today we're diving into one of those, and trust me, it's going to be a fun ride! We're tasked with figuring out the equation of a cubic polynomial function. Now, what makes this one special is that we know its zeros and a specific point it passes through. Let's break it down.

Understanding the Zeros and the Cubic Polynomial

First off, what exactly are zeros of a function? In simple terms, the zeros are the x-values where the function's output (the y-value) is zero. Graphically, these are the points where the graph of the function crosses or touches the x-axis. For a cubic polynomial function, which is a polynomial of degree three (meaning the highest power of x is 3), we expect to have up to three real zeros. In our case, we're given these zeros explicitly: they are at x = 2, x = 3, and x = 5. This means that when x is 2, 3, or 5, the value of our cubic polynomial function, let's call it $f(x)$, is 0. So, we can write this as $f(2) = 0$, $f(3) = 0$, and $f(5) = 0$.

Knowing the zeros of a polynomial is super useful because it directly relates to its factored form. If a polynomial has zeros at $r_1, r_2, ext{and } r_3$, then it can be written in the form $f(x) = a(x - r_1)(x - r_2)(x - r_3)$, where 'a' is a constant that scales the entire function. This 'a' is crucial because it determines the vertical stretch or compression of the graph and whether it opens upwards or downwards. Without this 'a', we could have infinitely many cubic functions with the same zeros but different shapes and positions. So, for our problem, since the zeros are 2, 3, and 5, our cubic polynomial function will look something like $f(x) = a(x - 2)(x - 3)(x - 5)$.

Now, the problem also gives us another piece of vital information: the graph of this function passes through the coordinate $(0, -5)$. This means that when the input value (x) is 0, the output value (y or $f(x)$) is -5. We can write this as $f(0) = -5$. This specific point is going to be our key to finding the value of the constant 'a'.

Putting It All Together: Solving for 'a'

We have our general form of the cubic polynomial: $f(x) = a(x - 2)(x - 3)(x - 5)$. And we know that $f(0) = -5$. Let's substitute x = 0 and $f(x) = -5$ into our equation:

$-5 = a(0 - 2)(0 - 3)(0 - 5)$

Now, let's simplify the terms inside the parentheses:

$-5 = a(-2)(-3)(-5)$

Let's multiply these numbers together:

$(-2) imes (-3) = 6$

$6 imes (-5) = -30$

So, our equation becomes:

$-5 = a(-30)$

To find 'a', we need to isolate it. We can do this by dividing both sides of the equation by -30:

a = rac{-5}{-30}

Simplifying this fraction, we get:

a = rac{1}{6}

So, the constant 'a' is rac{1}{6}. This means our cubic polynomial function is scaled by a factor of rac{1}{6}.

The Final Equation and the Y-Intercept

Now that we've found the value of 'a', we can write the complete equation for our cubic polynomial function. We just substitute a = rac{1}{6} back into our general form:

f(x) = rac{1}{6}(x - 2)(x - 3)(x - 5)

This is the equation of the cubic polynomial function that has zeros at (2,0), (3,0), and (5,0) and passes through the point (0, -5). You can expand this if you need the polynomial in standard form ($ax^3 + bx^2 + cx + d$), but often, the factored form is more useful, especially when dealing with zeros.

Let's quickly do the expansion just for kicks. We'll multiply the binomials first:

$(x - 2)(x - 3) = x^2 - 3x - 2x + 6 = x^2 - 5x + 6$

Now multiply this by $(x - 5)$:

$(x^2 - 5x + 6)(x - 5) = x(x^2 - 5x + 6) - 5(x^2 - 5x + 6)$

$= x^3 - 5x^2 + 6x - 5x^2 + 25x - 30$

Combine like terms:

$= x^3 - 10x^2 + 31x - 30$

So, the expression inside the parentheses is $x^3 - 10x^2 + 31x - 30$. Now, multiply the whole thing by rac{1}{6}:

f(x) = rac{1}{6}(x^3 - 10x^2 + 31x - 30)

f(x) = rac{1}{6}x^3 - rac{10}{6}x^2 + rac{31}{6}x - rac{30}{6}

Simplifying the fractions where possible:

f(x) = rac{1}{6}x^3 - rac{5}{3}x^2 + rac{31}{6}x - 5

This is the standard form of our cubic polynomial function.

What is the Y-intercept?

Now, let's talk about the y-intercept. The y-intercept is the point where the graph of the function crosses the y-axis. This happens when the x-value is 0. So, to find the y-intercept, we simply evaluate the function at $x = 0$. Hey, guess what? The problem already gave us this information! It stated that the graph passes through the coordinate $(0, -5)$.

This means the y-intercept is -5. We can see this directly from the standard form of our equation: f(x) = rac{1}{6}x^3 - rac{5}{3}x^2 + rac{31}{6}x - 5. The constant term in a polynomial in standard form ($ax^n + ... + d$) is always the y-intercept. In this case, the constant term is -5.

So, there you have it! We've successfully found the equation of the cubic polynomial function and identified its y-intercept using the given zeros and the point it passes through. Math puzzles like these are awesome because they show how different pieces of information connect to reveal the whole picture. Keep practicing, guys, and you'll be solving these like pros in no time!