Find 'c' In Y = X^2 + 5x + C Through Point A(-2, 5)
Hey guys, let's dive into some awesome algebra today! We've got this super cool function, y = x^2 + 5x + c, and we know it passes through a specific point, A(-2, 5). Our mission, should we choose to accept it (and we totally should!), is to find the value of 'c'. It sounds a bit like a mystery, right? But don't worry, with a little bit of algebraic magic, we'll solve this puzzle in no time. So, grab your calculators, your notebooks, and let's get this party started!
Understanding the Function and the Point
Alright, let's break down what we're dealing with here. We have the equation y = x^2 + 5x + c. This is a quadratic function, which means it's going to graph as a parabola. The 'x^2' term is the superstar that gives it that parabolic shape. The '+ 5x' term influences where the parabola sits horizontally and its steepness. Now, the 'c' term, that's our mystery variable. It's the y-intercept of the parabola – where it crosses the y-axis. But here's the kicker: we don't know what 'c' is yet. This is where our friend, point A(-2, 5), comes into play. When we say a function passes through a point, it means that when x is -2, y must be 5. Think of it like this: the coordinates of point A are like a secret password that the function has to satisfy. If we plug these x and y values into our function, the equation will hold true. This is the fundamental principle we'll use to unlock the value of 'c'. It's like having a key (the point) to open a locked box (the value of 'c'). So, the first step is always to understand what you're given. We have the function's general form and a specific point it must go through. This gives us a concrete relationship between x, y, and our unknown 'c'. Remember, in algebra, points on a graph are just pairs of (x, y) values that make the equation true. So, our point A(-2, 5) tells us that when x = -2, then y = 5. We're going to substitute these known values into our equation. This is the core idea of working with functions and points – using the knowns to find the unknowns. It's a beautiful synergy, really. The function defines a relationship, and the point confirms a specific instance of that relationship. Let's get ready to plug 'em in!
Plugging In the Values: The Key Step
Now for the really exciting part, guys! We're going to take our function y = x^2 + 5x + c and our point A(-2, 5) and combine them. Remember, the point A(-2, 5) means that when x = -2, then y = 5. So, we'll substitute -2 wherever we see x in the equation, and we'll substitute 5 wherever we see y. It's like a direct substitution! Let's write it out:
5 = (-2)^2 + 5(-2) + c*
See what we did there? We replaced 'y' with '5' and 'x' with '-2'. This single step transforms our equation from a general form with an unknown 'c' into a specific equation that only has 'c' as the unknown. This is the power of using a point that lies on the graph. It gives us a specific condition that the equation must meet. Now, we just need to simplify this equation and solve for 'c'. It's going to be a piece of cake, I promise!
- First, let's calculate (-2)^2: (-2) * (-2) = 4. Easy peasy.
- Next, let's calculate 5(-2):* 5 * -2 = -10. Still with me?
So, our equation now looks like this:
5 = 4 + (-10) + c
We're getting closer to finding our mystery 'c'!
Simplifying and Solving for 'c'
We've made it this far, awesome job! Now we have the equation 5 = 4 + (-10) + c. Our goal is to isolate 'c' on one side of the equation. Let's simplify the right side first:
4 + (-10) = 4 - 10 = -6
So, the equation becomes:
5 = -6 + c
Look at that! We're almost there. To get 'c' all by itself, we need to get rid of that '-6' on the right side. How do we do that? We do the opposite operation! Since it's '-6', we'll add '6' to both sides of the equation to keep it balanced. Remember, whatever you do to one side, you must do to the other side in algebra. It's like a scale – you have to keep it even!
5 + 6 = -6 + c + 6
Now, let's do the addition:
5 + 6 = 11
And on the right side:
-6 + c + 6 = c (because -6 and +6 cancel each other out)
So, we end up with:
11 = c
BOOM! We found it! The value of 'c' is 11.
The Final Equation and Verification
So, we've discovered that c = 11. This means our original function, y = x^2 + 5x + c, is actually y = x^2 + 5x + 11. How cool is that? We started with a mystery, and now we have the complete picture. To make sure we're not dreaming, let's quickly verify this. We can plug our point A(-2, 5) back into our new equation, y = x^2 + 5x + 11, and see if it holds true.
Let's substitute x = -2:
y = (-2)^2 + 5*(-2) + 11
Calculate step-by-step:
y = 4 + (-10) + 11
y = 4 - 10 + 11
y = -6 + 11
y = 5
And guess what? We got y = 5! This matches the y-coordinate of our point A(-2, 5). So, our solution is correct! Our function y = x^2 + 5x + 11 definitely passes through the point A(-2, 5). It's always a good idea to double-check your work in algebra, guys. It builds confidence and ensures accuracy. So, the value of 'c' is indeed 11. High fives all around!
Conclusion: The Power of Substitution
In this algebraic adventure, we successfully found the value of 'c' for the function y = x^2 + 5x + c by using the information that it passes through the point A(-2, 5). The core concept we utilized was substitution. By substituting the coordinates of the given point (x = -2, y = 5) into the function's equation, we transformed a problem with two unknowns (y and c) into an equation with just one unknown (c). We then simplified and solved this linear equation to find that c = 11. This method of substitution is a fundamental technique in algebra and is super useful for solving a wide range of problems, from finding unknown parameters in functions to solving systems of equations. Remember, when a graph passes through a specific point, those (x, y) coordinates must satisfy the equation of the graph. This seemingly simple idea is incredibly powerful for unlocking algebraic mysteries. So, the next time you see a function and a point it passes through, you know exactly what to do: plug those values in and solve for the unknown! Keep practicing, and you'll become an algebra whiz in no time. You guys got this!