Parabolic Vertex Solutions In Systems Of Equations

When Tom declared that the system of equations he was working with has two solutions, and one of them specifically landed on the vertex of the parabola, he set a pretty interesting mathematical challenge. Let's dive into what this means and the specific qualifications that must be met for Tom's thinking to be absolutely spot-on. We're talking about a system involving a parabola and a line, and the magic happens when they intersect in a particular way. The two equations we're looking at are:

Equation 1: $(x-3)^2 = y-4$ Equation 2: $y = -x + b$

For Tom's assertion to hold water, we need to unpack what it means for a solution to be at the vertex of the parabola, and how that interacts with the second equation, the line. The first equation, $(x-3)^2 = y-4$, is the standard form of a parabola that opens upwards. We can rewrite it as $y = (x-3)^2 + 4$. From this standard vertex form, it's crystal clear that the vertex of this parabola is located at the point $(3, 4)$. This is a fundamental property of parabolas in this form: the vertex is at $(h, k)$ when the equation is written as $y = a(x-h)^2 + k$. Here, $a=1$, $h=3$, and $k=4$, so our vertex is indeed $(3, 4)$.

Now, for one of the solutions to be at this vertex, the coordinates of the vertex $(3, 4)$ must satisfy both equations in the system. We already know it satisfies Equation 1 by definition of it being the vertex. So, the crucial part is that $(3, 4)$ must also satisfy Equation 2. If we substitute $x=3$ and $y=4$ into $y = -x + b$, we get $4 = -(3) + b$. Solving for $b$, we find $4 = -3 + b$, which means $b = 4 + 3$, so $b=7$. Therefore, a primary qualification for Tom's thinking to be correct is that the value of $b$ in the second equation must be 7. If $b$ is not 7, then the line $y = -x + b$ will not pass through the vertex $(3, 4)$, and thus $(3, 4)$ cannot be a solution to the system.

But wait, there's more! Tom also stated that the system has two solutions. This means the line $y = -x + b$ must intersect the parabola $y = (x-3)^2 + 4$ at exactly two distinct points. We've already established that one of these points is the vertex $(3, 4)$, which occurs when $b=7$. Now we need to confirm if, with $b=7$, there is indeed another intersection point. To find the intersection points, we set the expressions for $y$ equal to each other:

$(x-3)^2 + 4 = -x + 7$

Let's expand and simplify this equation to see how many solutions it yields:

$x^2 - 6x + 9 + 4 = -x + 7$ $x^2 - 6x + 13 = -x + 7$

Now, let's move all terms to one side to form a quadratic equation:

$x^2 - 6x + x + 13 - 7 = 0$ $x^2 - 5x + 6 = 0$

This is a quadratic equation in the form $ax^2 + bx + c = 0$, where $a=1$, $b=-5$, and $c=6$. To determine the number of solutions, we can look at the discriminant, $\Delta = b^2 - 4ac$. If $\Delta > 0$, there are two distinct real solutions. If $\Delta = 0$, there is exactly one real solution (a repeated root). If $\Delta < 0$, there are no real solutions.

Let's calculate the discriminant for our equation $x^2 - 5x + 6 = 0$:

$\Delta = (-5)^2 - 4(1)(6)$ $\Delta = 25 - 24$ $\Delta = 1$

Since the discriminant $\Delta = 1$, which is greater than 0, this quadratic equation has two distinct real solutions for $x$. This is fantastic news for Tom! It means that when $b=7$, the line $y = -x + 7$ intersects the parabola $y = (x-3)^2 + 4$ at two distinct points. We already know one of these points is the vertex $(3, 4)$ because we forced $b$ to be 7. Let's verify that $x=3$ is indeed a solution to $x^2 - 5x + 6 = 0$:

$(3)^2 - 5(3) + 6 = 9 - 15 + 6 = 0$. Yes, it is!

Now, let's find the other solution for $x$ by factoring the quadratic equation $x^2 - 5x + 6 = 0$. We are looking for two numbers that multiply to 6 and add up to -5. These numbers are -2 and -3. So, the equation factors as:

$(x-2)(x-3) = 0$

This gives us two solutions for $x$: $x-2 = 0 \implies x = 2$ and $x-3 = 0 \implies x = 3$. We already know that $x=3$ corresponds to the vertex solution. Let's find the corresponding $y$-value for $x=2$ using the line equation $y = -x + 7$ (since we've established $b=7$):

$y = -(2) + 7$ $y = 5$

So, the second intersection point is $(2, 5)$.

Therefore, for Tom's thinking to be correct, two primary qualifications must be met:

1. The value of $b$ must be 7. This ensures that the line passes through the vertex of the parabola, making the vertex one of the solutions to the system.
2. The discriminant of the resulting quadratic equation must be positive. This ensures that there is a second distinct intersection point between the line and the parabola, fulfilling the condition that the system has two solutions.

In summary, Tom's assertion is correct if and only if $b=7$. With $b=7$, the system yields the solutions $(3, 4)$ (the vertex) and $(2, 5)$, thereby satisfying both conditions: the system has two solutions, and one of them is the vertex of the parabola. It's a neat intersection of algebraic and geometric properties of these functions!

For further exploration into systems of equations and their graphical interpretations, you can visit the Khan Academy Mathematics section. They offer a comprehensive range of resources and explanations on these topics.