Unlock Quadratic Secrets: Find A+b In F(x)=(x-a)(x-b)

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Unlock Quadratic Secrets: Find a+b in f(x)=(x-a)(x-b)

Hey there, math enthusiasts and problem-solvers! Ever stared at a math problem and thought, "There has to be a trick to this?" Well, today we're diving deep into a super cool quadratic function problem involving the form f(x)=(xβˆ’a)(xβˆ’b){f(x)=(x-a)(x-b)}. We're given some intriguing clues: f(18)>0{f(18)>0}, f(21)<0{f(21)<0}, and f(24)>0{f(24)>0}. Our mission, should we choose to accept it, is to figure out the sum of a+b{a+b}. This isn't just about crunching numbers; it’s about understanding the behavior of parabolas and using their inherent symmetry to our advantage. If you're ready to uncover some neat algebraic insights and truly understand quadratic functions, stick around! This challenge might seem tricky at first, but once you grasp the fundamental properties of these functions, you'll see just how elegant the solution truly is. We’ll break down every step, making sure you not only get the answer but also understand why it works. So, let’s get started and unravel the mystery of these roots and their sum!

Decoding the Clues: What f(18)>0, f(21)<0, and f(24)>0 Really Tell Us

Alright, guys, let's break down these inequalities because they are packed with vital information about our quadratic function, f(x)=(xβˆ’a)(xβˆ’b){f(x)=(x-a)(x-b)}. First things first, remember that f(x)=(xβˆ’a)(xβˆ’b){f(x)=(x-a)(x-b)} represents a parabola that opens upwards. How do we know it opens upwards? Because if you multiply it out, the x2{x^2} term will have a positive coefficient (specifically, 1). This is super important because the direction a parabola opens dictates how its function values relate to its roots. The roots a{a} and b{b} are simply the points where the parabola crosses the x-axis, meaning where f(x)=0{f(x)=0}. Imagine that beautiful U-shaped curve; if it opens upwards, the function values are negative between the roots and positive outside the roots.

Now, let's look at our specific clues:

  • f(21)<0{f(21)<0}: This tells us that when x=21{x=21}, the function's value is negative. For an upward-opening parabola, a negative function value means that x=21{x=21} must be located between the roots a{a} and b{b}. So, we know that one root is smaller than 21, and the other is larger. Without loss of generality, let's assume a<b{a < b}, which means we have a<21<b{a < 21 < b}.

  • f(18)>0{f(18)>0}: This clue indicates that at x=18{x=18}, the function's value is positive. Since 18{18} is less than 21{21}, and 21{21} is already between our roots a{a} and b{b}, x=18{x=18} must be to the left of the smaller root a{a}. If 18{18} were between a{a} and b{b}, f(18){f(18)} would be negative. If 18{18} were to the right of b{b}, that would contradict 18<21<b{18 < 21 < b}. So, we can confidently say that 18<a{18 < a}.

  • f(24)>0{f(24)>0}: Similarly, at x=24{x=24}, the function's value is positive. Given that 24{24} is greater than 21{21}, and 21{21} is between a{a} and b{b}, x=24{x=24} must be to the right of the larger root b{b}. If 24{24} were between a{a} and b{b}, f(24){f(24)} would be negative. So, we know that b<24{b < 24}.

Putting all these pieces together, we've established a very specific ordering for our roots a{a} and b{b}: 18<a<21<b<24{18 < a < 21 < b < 24}. This interval analysis is absolutely crucial. It narrows down the possible locations for a{a} and b{b} significantly. Think of it like a scavenger hunt where each clue eliminates possibilities until you're left with a very clear map. We've just mapped out where our roots of a quadratic must lie on the number line, which is a fantastic start to solving this puzzle! This precise understanding of parabola behavior sets us up for the big reveal in the next section. We're not just solving for an answer; we're appreciating the inherent logic of these mathematical relationships.

The Symmetry Secret: Unveiling the Vertex of the Parabola

Now, here’s where the real magic of quadratic properties comes into play, guys. Every parabola has a special point called the vertex, which is either its absolute minimum or maximum. For our upward-opening parabola, the vertex is the lowest point. This vertex also lies on the axis of symmetry, an imaginary vertical line that cuts the parabola perfectly in half. For a quadratic function in the form f(x)=(xβˆ’a)(xβˆ’b){f(x)=(x-a)(x-b)}, the x-coordinate of this vertex (and thus the axis of symmetry) is always given by the average of its roots: xvertex=(a+b)/2{x_{vertex} = (a+b)/2}. Our ultimate goal is to find a+b{a+b}, so if we can pinpoint xvertex{x_{vertex}}, we're practically home free!

Let’s recall our clues: f(18)>0{f(18)>0}, f(21)<0{f(21)<0}, and f(24)>0{f(24)>0}. The key insight here isn't just in the signs, but in the relationship between the x-values. Notice that 21{21} is exactly in the middle of 18{18} and 24{24}. That is, (18+24)/2=42/2=21{(18+24)/2 = 42/2 = 21}. This equidistant spacing is a huge hint!

To really nail this down, let’s create a little helper function. Let's define a new function, h(x)=f(x)βˆ’f(21){h(x) = f(x) - f(21)}. Why do this? Because h(x){h(x)} is still a quadratic function with the exact same shape as f(x){f(x)} (it just shifted vertically). It also opens upwards because it shares the same leading coefficient of 1. What’s special about h(x){h(x)}?

  • First, let’s evaluate h(21){h(21)}: h(21)=f(21)βˆ’f(21)=0{h(21) = f(21) - f(21) = 0}. So, h(x){h(x)} crosses the x-axis at x=21{x=21}.

  • Next, let's look at h(18){h(18)}: We know f(18)>0{f(18)>0} and f(21)<0{f(21)<0}. If you take a positive number and subtract a negative number, you definitely get a positive number! So, h(18)=f(18)βˆ’f(21)>0{h(18) = f(18) - f(21) > 0}.

  • Finally, for h(24){h(24)}: Again, f(24)>0{f(24)>0} and f(21)<0{f(21)<0}. So, h(24)=f(24)βˆ’f(21)>0{h(24) = f(24) - f(21) > 0}.

Think about this carefully, folks. We have an upward-opening parabola h(x){h(x)} where h(21)=0{h(21)=0}, but for points to the left (like x=18{x=18}) and to the right (like x=24{x=24}) of 21{21}, the function values are positive. The only way an upward-opening parabola can have a value of zero at a point, with all other surrounding values being positive, is if that point is its absolute minimum – its vertex! If the vertex were anywhere else, say at xvβ‰ 21{x_v \neq 21}, then h(21){h(21)} would have to be positive because the lowest point would be at xv{x_v}. Therefore, x=21{x=21} must be the x-coordinate of the vertex for h(x){h(x)}.

Since h(x){h(x)} is just a vertical translation of f(x){f(x)}, they share the exact same axis of symmetry and the exact same x-coordinate for their vertex. This is a crucial piece of our problem-solving technique! This means the vertex of our original function, f(x){f(x)}, is also at x=21{x=21}. We’ve just found the central point of our parabola, which holds the key to the sum of roots!

The Grand Reveal: Calculating a+b

Alright, my clever friends, we've done all the hard work, and now it's time for the grand reveal! We've meticulously analyzed the clues, understood the behavior of quadratic functions, and most importantly, we uncovered the hidden symmetry that points directly to our answer. From the previous section, we established beyond a shadow of a doubt that the vertex of our parabola, f(x)=(xβˆ’a)(xβˆ’b){f(x)=(x-a)(x-b)}, is located at x=21{x=21}. This was thanks to our ingenious auxiliary function h(x){h(x)} which helped us pinpoint the minimum. Remember, the vertex is where the parabola hits its lowest point (since it opens upwards) and it’s also the center of its symmetry.

Now, let's put it all together. We know a fundamental property of a quadratic function in the form f(x)=Ax2+Bx+C{f(x)=Ax^2+Bx+C} (or factored as f(x)=(xβˆ’a)(xβˆ’b){f(x)=(x-a)(x-b)}) is that its axis of symmetry, and thus the x-coordinate of its vertex, is given by xvertex=βˆ’B2A{x_{vertex} = \frac{-B}{2A}}. In our specific form, f(x)=(xβˆ’a)(xβˆ’b){f(x)=(x-a)(x-b)} expands to f(x)=x2βˆ’(a+b)x+ab{f(x) = x^2 - (a+b)x + ab}. Comparing this to Ax2+Bx+C{Ax^2+Bx+C}, we see that A=1{A=1} and B=βˆ’(a+b){B=-(a+b)}. Plugging these into the vertex formula, we get:

xvertex=βˆ’(βˆ’(a+b))2(1){x_{vertex} = \frac{-(-(a+b))}{2(1)}}

xvertex=a+b2{x_{vertex} = \frac{a+b}{2}}

And just moments ago, we determined that our xvertex{x_{vertex}} is 21{21}! So, we can set up this simple algebraic calculation:

a+b2=21{\frac{a+b}{2} = 21}

To find the sum of a+b{a+b}, all we need to do is multiply both sides of the equation by 2:

a+b=21Γ—2{a+b = 21 \times 2}

a+b=42{a+b = 42}

There it is! The seemingly complex problem boils down to a single, elegant number. The beauty of this quadratic equation solution is its certainty. The conditions provided, combined with the inherent symmetry of parabolas, uniquely determine this sum. As a quick sanity check, think about the example we informally considered earlier: if a=19{a=19} and b=23{b=23}, then a+b=42{a+b=42}. Let's check our conditions for f(x)=(xβˆ’19)(xβˆ’23){f(x)=(x-19)(x-23)}:

  • f(18)=(18βˆ’19)(18βˆ’23)=(βˆ’1)(βˆ’5)=5{f(18) = (18-19)(18-23) = (-1)(-5) = 5}, which is >0{>0}. Perfect!
  • f(21)=(21βˆ’19)(21βˆ’23)=(2)(βˆ’2)=βˆ’4{f(21) = (21-19)(21-23) = (2)(-2) = -4}, which is <0{<0}. Perfect!
  • f(24)=(24βˆ’19)(24βˆ’23)=(5)(1)=5{f(24) = (24-19)(24-23) = (5)(1) = 5}, which is >0{>0}. Perfect!

This example perfectly fits all the initial conditions and yields a+b=42{a+b=42}, providing strong confidence in our mathematical proof. This problem isn't just about getting an answer; it’s about appreciating how a deep understanding of quadratic properties can reveal elegant solutions.

Beyond the Numbers: Why This Matters for Your Math Journey

So, we've cracked the code, found the elusive sum of roots, a+b=42{a+b=42}, and had a blast doing it (I hope!). But really, what’s the big takeaway from tackling a mathematical problem-solving challenge like this? It's not just about getting the right answer; it's about the journey and the skills you pick up along the way. This problem is a fantastic example of how understanding the visual and symmetrical properties of functions can lead to powerful algebraic solutions.

First, we reinforced our understanding of quadratic functions and parabolas. We saw how the sign of f(x){f(x)} tells us whether we are between or outside the roots when the parabola opens upwards. This interval analysis is a fundamental concept in pre-calculus and calculus, helping us understand where functions are positive or negative, which is crucial for sketching graphs and solving inequalities. This ability to interpret numerical conditions graphically is a superpower in math!

Second, we engaged in critical thinking by constructing an auxiliary function, h(x)=f(x)βˆ’f(21){h(x) = f(x) - f(21)}. This brilliant move transformed a problem with inequalities into one where a specific point became the obvious vertex. This technique of transforming a problem to simplify it or reveal hidden structures is incredibly valuable, not just in math but in all areas of problem-solving. It teaches us to look beyond the surface and manipulate given information creatively.

Third, and perhaps most importantly, we leveraged the power of symmetry. The fact that 21{21} was exactly midway between 18{18} and 24{24} was no coincidence; it was a carefully placed clue that, when combined with the properties of quadratic functions, led us directly to the axis of symmetry. Recognizing and utilizing symmetry is a cornerstone of advanced mathematics and physics. It helps simplify complex systems and uncover underlying order. Think about it: without understanding that symmetry meant the vertex was at x=21{x=21}, we might have floundered with complex inequalities.

This entire exercise strengthens your mathematical concepts and logical deduction. It shows that math isn't just about memorizing formulas; it's about understanding why those formulas work and how to apply them flexibly. Whether you're heading into higher-level math, pursuing a career in STEM, or just enjoy a good brain teaser, the skills honed here – careful analysis, strategic thinking, and appreciating mathematical elegance – will serve you well on your learning journey. Keep exploring, keep questioning, and you'll find that math holds endless fascinating secrets just waiting to be unlocked!