Hey guys! Let's dive into an interesting parabola problem. We're given that the vertex of a parabola is located at the point (1, -7), and this parabola crosses the x-axis somewhere between 6 and 7. Our mission? To figure out between which two negative integers this parabola also intersects the x-axis. Sounds like fun, right? Let's break it down!
Understanding Parabolas and Their Properties
When dealing with parabola problems, it's super important to first nail down the basics. A parabola, at its heart, is a U-shaped curve, and it's a key player in quadratic equations. The most basic form of a quadratic equation is y = ax² + bx + c, and when you graph this, you get a parabola. Now, the vertex? That's the turning point of the parabola – either the very bottom (if the parabola opens upwards) or the very top (if it opens downwards).
Think of the parabola's vertex as the heart of the curve. It’s the point where the parabola changes direction. If a in the equation y = ax² + bx + c is positive, the parabola opens upwards, making the vertex the minimum point. If a is negative, the parabola opens downwards, and the vertex is the maximum point. In our case, since the parabola intersects the x-axis, and the vertex is at (1, -7), which is below the x-axis, we know our parabola opens upwards. This means a is positive, and (1, -7) is the lowest point on our curve.
Another crucial concept is the axis of symmetry. Imagine drawing a vertical line right through the vertex; this line perfectly cuts the parabola into two mirror-image halves. The x-coordinate of the vertex gives us the equation for this line. So, in our scenario, the axis of symmetry is the vertical line x = 1. This symmetry is super helpful because it tells us that if we know one x-intercept (where the parabola crosses the x-axis), we can easily find the other by reflecting it across this line of symmetry. These x-intercepts, by the way, are also known as the roots or zeros of the quadratic equation – they're the x values that make y equal to zero.
Knowing the vertex and understanding the symmetry of a parabola can unlock a lot of information about its graph and equation. It allows us to predict where the parabola will go and how it will behave, which is exactly what we need to solve this problem.
Using the Vertex Form of a Quadratic Equation
Okay, so we know the vertex, and we need to find the other x-intercept. The vertex form of a quadratic equation is going to be our best friend here. This form is written as y = a(x - h)² + k, where (h, k) is the vertex of the parabola. Why is this so useful? Because we already know the vertex! In our problem, the vertex is (1, -7), so we can plug these values directly into the vertex form. This gives us y = a(x - 1)² - 7.
Notice that a is still there. This a is the same leading coefficient we talked about earlier, and it dictates how “wide” or “narrow” the parabola is, as well as which direction it opens. To find the exact value of a, we need a little more information. This is where the x-intercept comes into play. We know the parabola intersects the x-axis between 6 and 7. When the parabola intersects the x-axis, the y-coordinate is 0. So, we have a point (x, 0), where x is somewhere between 6 and 7. Let’s call this x-intercept x₁, so 6 < x₁ < 7.
We can use this information to solve for a. Plug y = 0 and x = x₁ into our vertex form equation: 0 = a(x₁ - 1)² - 7. Now, we need to isolate a. Add 7 to both sides to get 7 = a(x₁ - 1)². Then, divide both sides by (x₁ - 1)² to solve for a: a = 7 / (x₁ - 1)². This gives us an expression for a in terms of x₁. We don't need the exact value of x₁ to solve the problem; we just need this expression to help us find the other x-intercept.
This is a crucial step because it links the known x-intercept to the shape and direction of the parabola. By plugging in the vertex coordinates and using the given x-intercept, we've set up an equation that allows us to determine the leading coefficient a. This coefficient is vital for finding the other x-intercept, which will lead us to our final answer.
Finding the Second x-intercept Using Symmetry
Now, let's leverage the power of symmetry! We know the axis of symmetry is the vertical line x = 1. If one x-intercept, x₁, is somewhere between 6 and 7, the other x-intercept, let's call it x₂, will be the mirror image of x₁ across this line. Remember, the axis of symmetry cuts the parabola perfectly in half, so the distance from the axis of symmetry to x₁ is the same as the distance from the axis of symmetry to x₂.
To find x₂, we can use a simple formula based on the axis of symmetry. The midpoint between the two x-intercepts must lie on the axis of symmetry. So, the average of x₁ and x₂ must equal the x-coordinate of the vertex, which is 1. This gives us the equation: (x₁ + x₂) / 2 = 1. Multiply both sides by 2 to get x₁ + x₂ = 2. Now, solve for x₂: x₂ = 2 - x₁.
We know that 6 < x₁ < 7. To find the range for x₂, we'll use these inequalities. First, subtract x₁ from all parts of the equation x₂ = 2 - x₁: x₂ = 2 - x₁. Since we're dealing with inequalities, we need to be careful with the signs. If 6 < x₁ < 7, then -7 < -x₁ < -6. Now, add 2 to all parts of the inequality: 2 - 7 < 2 - x₁ < 2 - 6, which simplifies to -5 < x₂ < -4.
So, what does this tell us? It means that the other x-intercept, x₂, lies between -5 and -4. That’s it! We've found the two negative integers between which the parabola intersects the x-axis. The magic of symmetry, guys, it's pretty cool!
Conclusion: Putting It All Together
Let's recap the whole process. We started with the vertex of a parabola at (1, -7) and the knowledge that it intersects the x-axis between 6 and 7. Using the vertex form of a quadratic equation, y = a(x - h)² + k, we plugged in the vertex coordinates and found an expression for the leading coefficient a. We then used the symmetry of the parabola around the axis of symmetry x = 1 to find the second x-intercept. By calculating the mirror image of the given x-intercept across the axis of symmetry, we determined that the other x-intercept lies between -5 and -4.
Therefore, the correct answer is that the parabola intersects the x-axis between -5 and -4. Isn’t it amazing how much we can figure out about a parabola just by knowing its vertex and one other point? This problem showcases the beauty and power of quadratic equations and their graphical representation. Keep practicing, and you'll be a parabola pro in no time!
So, in the end, the answer is:
C. between -6 and -5
