Parabola Vertex In Quadrant I: Finding The Smallest 'a'

Hey guys! Let's dive into a cool algebra problem. We're tasked with finding the smallest integer value of 'a' that places the vertex of the parabola, defined by the equation y = ax² - 4x + a/2, in the first quadrant. This involves understanding parabolas, their vertices, and the coordinate plane. It's like a treasure hunt, but instead of gold, we're after the right value of 'a'!

Understanding the Problem and Key Concepts

Alright, so what does this all mean? First off, the first quadrant is that upper-right area of the coordinate plane where both x and y values are positive. Our goal is to make sure the vertex of our parabola – that special point where the curve changes direction – lands in this sweet spot. To do this, we need to manipulate the value of 'a' in the equation and figure out how it affects the parabola's position. Remember that the vertex form of a parabola is super useful here! It helps us visualize where the vertex actually sits.

The standard form of a quadratic equation is y = ax² + bx + c. In our case, we have y = ax² - 4x + a/2. The coefficient 'a' in front of the x² term is crucial; it determines whether the parabola opens upwards (if a > 0) or downwards (if a < 0). Also, the x-coordinate of the vertex can be found using the formula x = -b / 2a. Once we have the x-coordinate, we can plug it back into the original equation to get the y-coordinate of the vertex. This gives us the exact location of the vertex in the coordinate plane. Finding the vertex is the key to solving this problem; it’s like finding the 'X' on the treasure map!

For our parabola to have its vertex in the first quadrant, both the x and y coordinates of the vertex must be positive. We'll use this fact to set up inequalities and solve for 'a'. This is all about finding the right balance, like adjusting the sails on a boat to catch the wind. Let's get our hands dirty and solve this thing!

Finding the Vertex Coordinates

Okay, let's get down to business and figure out the vertex's coordinates. Remember, our equation is y = ax² - 4x + a/2. To find the x-coordinate of the vertex, we'll use the formula x = -b / 2a. Here, b = -4, so:

x = -(-4) / 2a = 4 / 2a = 2 / a

Now we know the x-coordinate of the vertex is 2/a. For the vertex to be in the first quadrant, 2/a must be positive. This means that if a is positive, the x-coordinate will be positive. Great!

Next, we need to find the y-coordinate. We'll substitute x = 2/a back into the original equation:

y = a(2/a)² - 4(2/a) + a/2 y = a(4/a²) - 8/a + a/2 y = 4/a - 8/a + a/2 y = -4/a + a/2

So the y-coordinate of the vertex is -4/a + a/2. For the vertex to be in the first quadrant, this value must also be positive. This is where the fun starts! We now have two conditions:

1. a must be positive (from the x-coordinate requirement)
2. -4/a + a/2 > 0 (from the y-coordinate requirement)

Let's solve that second inequality to find out more about 'a'. This is the part where we bring all of our skills into play. Pay close attention, because this is the heart of the problem!

Solving the Inequality and Finding 'a'

Alright, let's tackle that inequality: -4/a + a/2 > 0. We need to isolate 'a' to figure out the range of values that satisfy this condition. First, let's get rid of the fraction by multiplying through by 2a. Remember, we know a must be positive. If we multiply by a negative number, we will need to flip the inequality sign. So, multiplying by 2a, we get:

2a(-4/a + a/2) > 2a(0)

-8 + a² > 0

Now, let's rearrange this to a² > 8. To solve this, we can take the square root of both sides. However, we must consider both the positive and negative roots. So, we get a > √8 or a < -√8. Since √8 is approximately 2.83, this tells us a > 2.83 or a < -2.83.

But, remember that 'a' also has to be positive for the x-coordinate to be positive, and we know a is already greater than zero, so we must take into account the intersection of our two conditions. It means a > √8 which is approximately 2.83. Since we're looking for the smallest integer value of 'a', we must find the smallest whole number that is greater than 2.83. Therefore, the smallest integer value for 'a' is 3. That is the answer!

We've successfully navigated the inequalities and have found the perfect value for 'a' to place the vertex in the first quadrant. Congratulations on solving this problem. You should be proud of yourself!

Final Answer and Conclusion

So, after all that number crunching, we've found that the smallest integer value for 'a' that places the vertex of the parabola in the first quadrant is 3. Pretty cool, right?

We’ve explored the characteristics of parabolas, the coordinate plane, and how changing a coefficient can alter the graph. We used the vertex formula, solved inequalities, and made sure both the x and y coordinates of the vertex were positive. Understanding these concepts is crucial for mastering algebra and solving more complex problems. Keep practicing, and you'll become a problem-solving wizard in no time!

Remember that the key is to break down the problem into smaller, more manageable parts. Always start with the knowns (the equation and the desired quadrant) and use the appropriate formulas and concepts to find the unknowns (the vertex and 'a').

Keep up the excellent work, and happy problem-solving, guys!