Determine The Quadratic Function Whose Graph Is Given Below

To determine thequadratic function whose graph is given below, you must first decode the visual clues that the parabola reveals about its equation. A quadratic function has the standard form

[ f(x)=ax^{2}+bx+c ]

where a, b, and c control the shape, direction, and position of the curve. By extracting key characteristics—such as the vertex, axis of symmetry, x‑intercepts, and a single additional point—you can set up a system of equations that uniquely identifies the coefficients. This process blends algebraic manipulation with geometric interpretation, turning a plotted curve into a precise mathematical expression.

Understanding the Graph’s Core Features

Vertex and Direction

The vertex is the highest or lowest point of the parabola, depending on whether it opens upward (a > 0) or downward (a < 0). In the provided graph, the vertex appears at ((-2,,3)). This point satisfies the equation [ f(-2)=a(-2)^{2}+b(-2)+c=3]

The axis of symmetry runs vertically through the vertex, giving the line (x=-2). Knowing the axis helps locate symmetric points on either side of the curve.

Intercepts

The x‑intercepts (roots) are where the graph crosses the x‑axis, i.e., where (f(x)=0). From the picture, the parabola meets the x‑axis at (x=1) and (x=5). These zeros imply that the factored form can be written as

[ f(x)=a(x-1)(x-5) ]

The y‑intercept occurs at (x=0); the graph shows the curve passing through ((0,-4)). Substituting this point yields another equation involving a, b, and c.

Solving for the Coefficients

Setting Up the System

Using the three identified points, we obtain three equations:

1. From the vertex ((-2,3)): (4a-2b+c=3)
2. From the y‑intercept ((0,-4)): (c=-4)
3. From one root, say (x=1): (a(1)^{2}+b(1)+c=0)

With (c=-4), the first equation simplifies to (4a-2b-4=3) → (4a-2b=7).
The third equation becomes (a+b-4=0) → (a+b=4).

Now solve the linear system:

[\begin{cases} 4a-2b=7\ a+b=4 \end{cases} ]

Multiply the second equation by 2: (2a+2b=8). Adding to the first gives (6a=15) → (a=\dfrac{5}{2}). Substituting back, (b=4-a=4-\dfrac{5}{2}=\dfrac{3}{2}).

Writing the Final Function

Insert the coefficients into the standard form:

[ \boxed{f(x)=\dfrac{5}{2}x^{2}+\dfrac{3}{2}x-4} ]

If you prefer the factored version, use the root information and the calculated a:

[ f(x)=\frac{5}{2}(x-1)(x-5) ]

Both representations describe the same parabola.

Step‑by‑Step Procedure to Determine the Quadratic Function Whose Graph Is Given Below

1. Locate the vertex – read its coordinates ((h,k)).
2. Identify the direction – upward if the parabola opens like a smile, downward if it opens like a frown.
3. Find the x‑intercepts – note any visible roots; they provide factors of the equation. 4. Read the y‑intercept – the point where (x=0).
4. Choose a third distinct point – often a convenient point on the curve (e.g., a second root or a symmetric point).
5. Set up equations – substitute each point into (ax^{2}+bx+c) to generate a system.
6. Solve the system – use substitution or elimination to find a, b, and c.
7. Write the final expression – present the function in either standard or factored form, verifying that it matches all observed features.

Common Pitfalls and How to Avoid Them

• Misreading the vertex – ensure you capture the exact coordinates; a small shift changes the entire solution. - Assuming integer coefficients – the coefficients may be fractions or irrationals; keep calculations exact.
• Ignoring symmetry – the axis of symmetry can simplify calculations; use it to find missing points quickly.
• Overlooking a third point – with only two points, infinitely many quadratics fit; a third non‑collinear point is essential.
• Algebraic errors – double‑check each substitution and arithmetic step, especially when handling negatives.

Frequently Asked Questions (FAQ)

Q1: Can a quadratic function have more than two x‑intercepts?
A: No. A parabola can intersect the x‑axis at most twice, corresponding to its two real roots. Additional intercepts would imply a higher‑degree polynomial.

Q2: What if the graph shows only one x‑intercept?
A: A single x‑intercept indicates a repeated root (the vertex lies on the x‑axis). The factored form becomes (f(x)=a(x-h)^{2}).

Q3: How do I determine a without a third point?
A: If the vertex form is used, (f(x)=a(x-h)^{2}+k), the value of a can be inferred from the curvature’s steepness, but a precise numeric value still requires an additional point.

Q4: Is the vertex always the maximum or minimum?
A: Yes. For a > 0 the vertex is a minimum; for a < 0 it is a maximum. The sign of a dictates the direction of opening.

Q5: Can I use technology to verify my result?
A: Graphing calculators or software can plot the derived equation to confirm that it matches the original curve’s shape and key points.

Conclusion

By systematically extracting the vertex, intercepts, and a supplemental point from the plotted parabola, you can construct a reliable system of equations that yields the exact quadratic function. This method not only reinforces algebraic techniques such as substitution and factoring but also deepens geometric intuition about how algebraic expressions manifest as curves. Mastering the process of determine the quadratic function whose graph is given below equips students with a powerful tool for translating visual data into precise mathematical models, a skill that proves invaluable across physics,

…acrossphysics, engineering, economics, and data‑science. In each of these fields the ability to translate a visual pattern into an explicit algebraic rule enables prediction, optimization, and deeper insight.

Putting the pieces together
When a parabola is presented on a coordinate grid, the first step is to locate its vertex ((h,k)). From there the axis of symmetry can be drawn, offering a quick way to generate additional points that are mirror images across the line (x=h). Selecting any point that is not on the axis — say a point where the curve crosses a grid line — provides the third equation needed to solve for the coefficient (a). Substituting ((h,k)) into the vertex form yields

[f(x)=a(x-h)^{2}+k, ]

and solving for (a) using the third point gives the exact value of the leading coefficient. Expanding the expression, if desired, produces the standard form

[ f(x)=ax^{2}+bx+c, ]

where (b) and (c) emerge naturally from the algebraic manipulation. The resulting function reproduces every observed feature: the direction of opening, the exact location of the vertex, the x‑ and y‑intercepts, and the symmetry of the curve.

Beyond the basics
In more complex scenarios the graph may be given only partially — perhaps a segment of the curve or a set of plotted points that do not include the vertex explicitly. In such cases, fitting a quadratic through three non‑collinear points using a system of equations (e.g., solving for (a,b,c) directly) yields the same unique parabola, provided the points are not collinear. When the data are noisy or derived from experimental measurements, a regression approach can be employed to find the best‑fit quadratic, acknowledging that the exact curve may be approximated rather than perfectly captured.

Real‑world illustration
Consider a projectile launched from ground level with an initial velocity of (20\ \text{m/s}) at a (45^{\circ}) angle. Neglecting air resistance, its trajectory follows a parabola described by

[ y = -\frac{g}{2v_{0}^{2}\cos^{2}\theta},x^{2}+x\tan\theta, ]

where (g\approx9.8\ \text{m/s}^{2}) and (\theta=45^{\circ}). By extracting the vertex ((h,k)=\bigl(\frac{v_{0}^{2}\sin2\theta}{2g},\frac{v_{0}^{2}\sin^{2}\theta}{2g}\bigr)) from the motion equations, one can reconstruct the same quadratic directly from the plotted path, confirming that the analytical model matches the graphical representation.

Final take‑away
The process of determining the quadratic function whose graph is given below is more than a mechanical exercise; it is a bridge between visual intuition and symbolic precision. By systematically identifying key geometric attributes, constructing a minimal set of equations, and solving for the coefficients, students and professionals alike gain a reliable method for converting graphical information into a concrete mathematical model. This skill not only reinforces core algebraic techniques but also cultivates a mindset that views data — whether plotted on paper or displayed on a screen — as a gateway to deeper analytical understanding.