Quadratic Equation From Vertex And Point

Hey guys, ever needed to nail down the exact equation of a parabola when you've only got a vertex and another point? It sounds tricky, but trust me, it's totally doable and actually pretty cool once you get the hang of it. We're talking about finding that specific $y = ax^2 + bx + c$ equation, and knowing the vertex $(h, k)$ and a random point $(x, y)$ on the parabola is your golden ticket. This skill is super useful in all sorts of math stuff, from physics problems to graphing functions, so let's dive in and figure out how to solve this. We'll break down the steps, explain the logic, and by the end, you'll be a quadratic equation pro, ready to tackle any problem thrown your way. Get ready to flex those math muscles!

Understanding the Vertex Form of a Quadratic

Alright, before we jump into solving our specific problem, let's get reacquainted with the vertex form of a quadratic equation. It's not the $y = ax^2 + bx + c$ form we're aiming for, but it's our secret weapon to get there. The vertex form is written as $y = a(x - h)^2 + k$. See that? The $h$ and $k$ in this equation are directly the coordinates of your vertex. Pretty neat, right? This form is awesome because it immediately tells you where the parabola's turning point is. Once you have the vertex $(h, k)$, you can plug those values right into this form. So, if our vertex is $(-2, 4)$, we'd plug in $h = -2$ and $k = 4$. This gives us $y = a(x - (-2))^2 + 4$, which simplifies to $y = a(x + 2)^2 + 4$. Now, we've used the vertex information! We have an equation with just one unknown: the 'a' value. This 'a' value is crucial because it controls the parabola's width and whether it opens upwards (if 'a' is positive) or downwards (if 'a' is negative). It dictates the parabola's stretch or compression. Without knowing 'a', we'd just have a family of parabolas all sharing the same vertex. Our next step is to figure out that specific 'a' that makes our parabola unique. This is where the other piece of information comes in handy – the extra point the parabola passes through. By plugging the coordinates of this known point into our vertex form equation, we can create a solvable equation for 'a'. It’s like having two pieces of a puzzle, and we’re using each piece to reveal the full picture of our quadratic.

Using the Given Point to Find 'a'

So, we've got our vertex form equation, $y = a(x + 2)^2 + 4$, thanks to our vertex $(-2, 4)$. Now, we need to find that elusive 'a' value. This is where the second piece of information comes into play: the point $(3, 9)$ that our parabola passes through. Since this point is on the parabola, its x and y coordinates must satisfy the equation. This means we can substitute $x = 3$ and $y = 9$ directly into our current equation. Let's do it: $9 = a(3 + 2)^2 + 4$. Now, we just have a simple algebraic equation with only 'a' as our variable. Let's solve it step-by-step, guys. First, simplify inside the parentheses: $9 = a(5)^2 + 4$. Squaring the 5 gives us $9 = a(25) + 4$. To isolate the term with 'a', we subtract 4 from both sides: $9 - 4 = 25a$, which simplifies to $5 = 25a$. Finally, to find 'a', we divide both sides by 25: a = rac{5}{25}, which simplifies to a = rac{1}{5}. Boom! We've found our 'a' value. This tells us our parabola opens upwards (since $a$ is positive) and is somewhat wider than a standard parabola. With 'a' in hand, we're super close to getting our final equation in the $y = ax^2 + bx + c$ form. We now have all the components needed to construct the complete vertex form of our specific quadratic equation. It’s like having all the ingredients ready to bake the perfect quadratic cake!

Converting to Standard Form ($y = ax^2 + bx + c$)

We've done the heavy lifting, guys! We figured out that a = rac{1}{5}, and we know our vertex form equation is y = rac{1}{5}(x + 2)^2 + 4. The final step is to convert this into the standard form, $y = ax^2 + bx + c$. This just involves a bit of algebraic expansion and simplification. First, we need to expand the $(x + 2)^2$ part. Remember, $(x + 2)^2$ means $(x + 2)(x + 2)$. Using the FOIL method (First, Outer, Inner, Last) or just distributing, we get: $x imes x = x^2$, $x imes 2 = 2x$, $2 imes x = 2x$, and $2 imes 2 = 4$. Combine the like terms ($2x + 2x$), and we get $x^2 + 4x + 4$. So, our equation now looks like: y = rac{1}{5}(x^2 + 4x + 4) + 4. The next step is to distribute that rac{1}{5} to each term inside the parentheses: rac{1}{5} imes x^2 = rac{1}{5}x^2, rac{1}{5} imes 4x = rac{4}{5}x, and rac{1}{5} imes 4 = rac{4}{5}. So, the equation becomes: y = rac{1}{5}x^2 + rac{4}{5}x + rac{4}{5} + 4. The last part is to combine the constant terms, rac{4}{5} and $4$. To add them, we need a common denominator. Since 4 = rac{20}{5}, we have rac{4}{5} + rac{20}{5} = rac{24}{5}. And voilà! We have our final equation in standard form: y = rac{1}{5}x^2 + rac{4}{5}x + rac{24}{5}. This is the specific quadratic equation that has a vertex at $(-2, 4)$ and passes through the point $(3, 9)$. You've successfully transformed the information into the desired format. Pretty awesome, right? You've mastered taking vertex and point information and turning it into the standard quadratic equation.

Verification and Conclusion

Now that we've arrived at our final equation, y = rac{1}{5}x^2 + rac{4}{5}x + rac{24}{5}, it's always a good idea to do a quick verification to make sure we didn't mess up anywhere along the way. Remember, we were given that the vertex is $(-2, 4)$ and a point on the parabola is $(3, 9)$. Let's first check if the vertex coordinates satisfy the equation. When $x = -2$, $y$ should be $4$. Plugging $x = -2$ into our equation: y = rac{1}{5}(-2)^2 + rac{4}{5}(-2) + rac{24}{5}. Calculate the terms: $(-2)^2 = 4$, so rac{1}{5}(4) = rac{4}{5}. Then rac{4}{5}(-2) = -rac{8}{5}. So, y = rac{4}{5} - rac{8}{5} + rac{24}{5}. Combine the numerators: $4 - 8 + 24 = 20$. So, y = rac{20}{5}, which simplifies to $y = 4$. Perfect! The vertex coordinates check out. Now, let's check the other point, $(3, 9)$. When $x = 3$, $y$ should be $9$. Plugging $x = 3$ into our equation: y = rac{1}{5}(3)^2 + rac{4}{5}(3) + rac{24}{5}. Calculate the terms: $(3)^2 = 9$, so rac{1}{5}(9) = rac{9}{5}. Then rac{4}{5}(3) = rac{12}{5}. So, y = rac{9}{5} + rac{12}{5} + rac{24}{5}. Combine the numerators: $9 + 12 + 24 = 45$. So, y = rac{45}{5}, which simplifies to $y = 9$. Excellent! This point also checks out. Both the vertex and the given point satisfy our derived equation. This confirms that y = rac{1}{5}x^2 + rac{4}{5}x + rac{24}{5} is indeed the correct quadratic equation for the given conditions. So, to wrap things up, remember the process: start with the vertex form $y = a(x - h)^2 + k$, plug in your vertex $(h, k)$, use the given point $(x, y)$ to solve for 'a', and finally, expand and simplify to get the standard form $y = ax^2 + bx + c$. You guys nailed it! Mastering this method gives you a powerful tool for analyzing and defining parabolic functions. Keep practicing, and you'll be finding quadratic equations like a pro in no time!