Graphing Quadratic Functions: Domain & Range Explained

Hey guys! Today, we're diving deep into the awesome world of quadratic functions and how to understand their graphs, specifically focusing on the statements about the function $f(x)= 2 x^2-x-6$. We'll be looking at its domain and range, and figuring out which statements about it are actually true. It's super important to nail down these concepts because they're fundamental to understanding how functions behave. Think of the domain as all the possible x values you can plug into the function, and the range as all the possible y (or f(x)) values you can get out. For quadratic functions, these are usually pretty predictable once you know what to look for. We'll break down each option so you can see exactly why it's right or wrong. Get ready to become a quadratic function graphing pro!

Understanding Domain and Range

Let's kick things off by getting a solid grip on what domain and range actually mean in the context of functions, especially for our friend $f(x)= 2 x^2-x-6$. The domain is essentially the set of all possible input values (the x-values) that a function can accept without breaking any math rules. Think of it as the 'allowed' x's. For most polynomial functions, like our quadratic one here, the domain is all real numbers. This means you can plug in any real number for x, and the function will give you a valid output. There are no restrictions like dividing by zero or taking the square root of a negative number that we need to worry about with standard polynomials. So, when you see statements about restricted domains, especially those involving inequalities like $x eq ext{something}$ or $x ext{ less than or greater than something}$, you should always pause and check if that restriction actually applies to the specific function you're looking at. Our function $f(x)= 2 x^2-x-6$ is a simple quadratic, so we expect its domain to be all real numbers unless there's something really unusual going on, which there isn't! Now, the range is the set of all possible output values (the y-values or f(x)-values) that the function can produce. This is where things can get a little more interesting, especially with quadratic functions. The range depends heavily on the direction the parabola opens and its vertex (the lowest or highest point). If a parabola opens upwards, its range will start at the y-coordinate of the vertex and go up to infinity. If it opens downwards, its range will start at negative infinity and go up to the y-coordinate of the vertex. So, the statement 'The range of the function is all real numbers' is only true for linear functions or functions with an odd degree that cover the entire y-axis. For our quadratic function, we'll need to figure out if this statement holds true. Keep these definitions in mind as we dissect the options provided for $f(x)= 2 x^2-x-6$!

Analyzing the Function $f(x)= 2 x^2-x-6$

Alright guys, let's get down to business with our specific function: $f(x)= 2 x^2-x-6$. This is a quadratic function because the highest power of x is 2. The general form of a quadratic function is $ax^2 + bx + c$. In our case, $a=2$, $b=-1$, and $c=-6$. The coefficient '$a$' (which is 2 here) is super important because it tells us the direction the parabola opens. Since '$a$' is positive ($a > 0$), the parabola opens upwards. This is a crucial piece of information because it directly impacts the range of the function. An upward-opening parabola has a minimum y-value, which is the y-coordinate of its vertex, and then it extends infinitely upwards. This means the range will not be all real numbers; it will be bounded from below. The domain, as we discussed, is typically all real numbers for any polynomial function, including quadratics. There are no values of x that will cause mathematical problems in $f(x)= 2 x^2-x-6$. You can plug in any positive number, negative number, or zero, and you'll get a perfectly valid real number as the output. So, any statement that tries to restrict the domain of this specific function to only certain x-values (like $x eq ext{a specific number}$, or $x ext{ greater than or equal to something}$) is likely incorrect. To find the exact range, we need to find the vertex of the parabola. The x-coordinate of the vertex is given by the formula $x = -b / (2a)$. For our function, this is $x = -(-1) / (2 imes 2) = 1 / 4$. Now, to find the y-coordinate of the vertex, we plug this x-value back into our function: $f(1/4) = 2(1/4)^2 - (1/4) - 6$. Let's calculate that: $f(1/4) = 2(1/16) - 1/4 - 6 = 1/8 - 2/8 - 48/8 = (1 - 2 - 48) / 8 = -49/8$. So, the vertex of our parabola is at the point $(1/4, -49/8)$. Since the parabola opens upwards, the minimum y-value is $-49/8$. Therefore, the range of the function is all real numbers greater than or equal to $-49/8$, which can be written as $[ -49/8, eq) $ or $f(x) eq -49/8$. Understanding this vertex and the direction of opening is key to accurately describing the graph's properties. This detailed analysis helps us to critically evaluate the given statements about the function's graph.

Evaluating Statement A: Domain Restriction

Let's tackle the first statement, guys: "A. The domain of the function is $\{x vert x geq \frac{1}{4}\}$." This statement is claiming that the only x-values that are allowed as input for our function $f(x)= 2 x^2-x-6$ are those that are greater than or equal to $1/4$. Now, remember what we learned about the domain of quadratic functions. Quadratic functions, especially those in the standard form $ax^2+bx+c$ with no denominators or square roots involving x, have a domain that encompasses all real numbers. This means you can substitute any real number for x into the function, and it will produce a valid output. There are no inherent mathematical restrictions that limit the x-values for $f(x)= 2 x^2-x-6$. The value $1/4$ that appears here is significant because it's the x-coordinate of the vertex of the parabola. However, the x-coordinate of the vertex only defines the minimum or maximum x-value if the function itself was defined with a restricted domain, which is not the case here. For example, if the problem had stated something like 'Consider the function $g(x) = 2x^2 - x - 6$ for $x geq 1/4$', then statement A might be relevant to that specific scenario. But as it stands, for the function $f(x)= 2 x^2-x-6$ with no explicit domain restrictions given, the domain is indeed all real numbers. Therefore, statement A, which imposes a restriction ($x geq 1/4$) on the domain, is false. It incorrectly limits the set of possible inputs for the function. We can plug in $x=0$, $x=-1$, $x=100$, or any other real number, and the function will work perfectly fine. The graph of $f(x)= 2 x^2-x-6$ extends infinitely to the left and right, covering all possible x-values. It's essential to distinguish between the natural domain of a function (all possible inputs) and a domain that might be artificially restricted by the problem statement or context. In this case, the natural domain is all real numbers, and statement A contradicts this fundamental property.

Evaluating Statement B: Range as All Real Numbers

Next up, let's analyze statement B: "B. The range of the function is all real numbers." This statement claims that the output values (y-values or $f(x)$-values) of our function $f(x)= 2 x^2-x-6$ can be any real number. We've already done some heavy lifting in understanding the implications of the coefficient '$a$' and the vertex. Remember, '$a$' is 2, which is positive ($a > 0$). This means our parabola opens upwards. When a parabola opens upwards, it has a lowest point, which is its vertex. The y-coordinate of this vertex represents the minimum value that the function can ever output. Any value below this minimum y-value is impossible for the function to produce. We calculated the vertex earlier and found it to be at $(1/4, -49/8)$. This means the minimum y-value, or the minimum output of the function, is $-49/8$. So, the range of $f(x)= 2 x^2-x-6$ is actually all real numbers greater than or equal to $-49/8$. This is typically written in interval notation as $[-49/8, eq) $ or using set notation as $\{y vert y geq -49/8\}$. Since the function cannot produce any y-values less than $-49/8$, the statement that the range is all real numbers is false. This statement would only be true for functions like linear functions with a non-zero slope, or cubic functions (and other odd-degree polynomials) that, due to their end behavior, cover the entire vertical axis. For quadratics with $a eq 0$, the range is always a half-line (either bounded below or bounded above), never the entire set of real numbers. Therefore, statement B is incorrect because it fails to acknowledge the minimum output value dictated by the upward-opening parabola and its vertex.

Evaluating Statement C: The Correct Range

Now, let's move on to statement C, which likely presents the correct description of the range: "C. The range of the function is $\{y vert y geq -49/8\}$." This statement asserts that the output values (y-values or $f(x)$-values) of our function $f(x)= 2 x^2-x-6$ are all real numbers greater than or equal to $-49/8$. Let's confirm this based on our previous analysis. We identified that $f(x)= 2 x^2-x-6$ is a quadratic function with $a=2$. Since $a > 0$, the parabola opens upwards. This upward direction means the function has a minimum value. This minimum value occurs at the vertex of the parabola. We calculated the x-coordinate of the vertex using the formula $x = -b/(2a)$, which gave us $x = -(-1)/(2 imes 2) = 1/4$. Then, we substituted this x-value back into the function to find the minimum y-coordinate: $f(1/4) = 2(1/4)^2 - (1/4) - 6 = 2(1/16) - 1/4 - 6 = 1/8 - 2/8 - 48/8 = -49/8$. So, the vertex is indeed at $(1/4, -49/8)$. Because the parabola opens upwards, the lowest y-value the function can ever achieve is $-49/8$. All other possible output values will be greater than this minimum. Therefore, the range of the function consists of all real numbers that are greater than or equal to $-49/8$. This can be precisely expressed in set notation as $\{y vert y geq -49/8\}$. This statement accurately reflects the behavior of an upward-opening parabola and its minimum output. Thus, statement C is true. It correctly identifies the boundary and direction of the function's output values.

Evaluating Statement D: Axis of Symmetry

Let's examine statement D: "D. The axis of symmetry of the graph is the line $x = 1/4$." The axis of symmetry is a vertical line that passes through the vertex of a parabola, dividing it into two mirror-image halves. For any quadratic function in the form $f(x) = ax^2 + bx + c$, the equation of the axis of symmetry is given by $x = -b/(2a)$. We've already used this formula to find the x-coordinate of the vertex of our function $f(x)= 2 x^2-x-6$. Let's plug in the values $a=2$ and $b=-1$ again: $x = -(-1) / (2 imes 2) = 1/4$. So, the axis of symmetry for the graph of $f(x)= 2 x^2-x-6$ is indeed the vertical line $x = 1/4$. This line perfectly bisects the parabola, meaning that for any distance you move horizontally away from this line, the function's value will be the same on both sides. This is a fundamental property of parabolas. Since our calculation directly matches the statement, statement D is true. It correctly identifies the line of symmetry for this specific quadratic function's graph.

Evaluating Statement E: Y-intercept

Finally, let's look at statement E: "E. The y-intercept of the graph is $(0, -3)$." The y-intercept is the point where the graph crosses the y-axis. This occurs when the x-value is zero. To find the y-intercept of any function, you simply substitute $x=0$ into the function's equation. For $f(x)= 2 x^2-x-6$, let's plug in $x=0$: $f(0) = 2(0)^2 - (0) - 6$. Calculating this gives us: $f(0) = 2(0) - 0 - 6 = 0 - 0 - 6 = -6$. So, the y-intercept is the point $(0, -6)$. The statement claims the y-intercept is $(0, -3)$. Since our calculated y-intercept is $(0, -6)$, statement E is false. It's a common mistake to miscalculate the constant term or confuse it with other parts of the function, but in $ax^2+bx+c$, the y-intercept is always at $(0, c)$. In our case, $c=-6$, so the y-intercept is $(0, -6)$, not $(0, -3)$.

Conclusion: Identifying the True Statements

After meticulously analyzing each statement regarding the function $f(x)= 2 x^2-x-6$, we can now confidently determine which ones are true. We found that:

• Statement A (Domain is $x geq 1/4$) is false because the domain of this quadratic function is all real numbers.
• Statement B (Range is all real numbers) is false because the upward-opening parabola has a minimum y-value.
• Statement C (Range is $y geq -49/8$) is true because it accurately describes the minimum output of the upward-opening parabola.
• Statement D (Axis of symmetry is $x = 1/4$) is true because it correctly identifies the vertical line passing through the vertex.
• Statement E (y-intercept is $(0, -3)$) is false because the y-intercept is actually $(0, -6)$.

Therefore, the two true statements about the graph of the function $f(x)= 2 x^2-x-6$ are C and D. It's super rewarding when you can break down a problem like this and pinpoint the correct answers with confidence. Keep practicing, and you'll master these concepts in no time!