The equation of a quadratic function describes a parabola, which is a U-shaped graph that can open upward or downward. Finding the equation of a quadratic function means determining the specific formula that matches a given graph, set of points, vertex, intercepts, or real-world situation. Quadratic functions are usually written in forms such as standard form, vertex form, or intercept form, and the best method depends on what information is given.
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Introduction to Quadratic Functions
A quadratic function is a polynomial function of degree 2. Its graph is called a parabola, and it has the general shape of a cup or an upside-down cup. The simplest quadratic function is:
[ f(x)=x^2 ]
This is called the parent function. Other quadratic functions are created by shifting, stretching, compressing, or reflecting this basic graph Turns out it matters..
The most common forms of a quadratic equation are:
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Standard form [ f(x)=ax^2+bx+c ]
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Vertex form [ f(x)=a(x-h)^2+k ]
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Intercept form [ f(x)=a(x-p)(x-q) ]
In each form, the value of a controls the direction and width of the parabola. On the flip side, if (a>0), the parabola opens upward. If (a<0), it opens downward.
Standard Form of a Quadratic Function
The standard form of a quadratic function is:
[ f(x)=ax^2+bx+c ]
In this form:
- (a), (b), and (c) are constants.
- (a\neq 0), because if (a=0), the function is no longer quadratic.
- The y-intercept is (c).
- The axis of symmetry is: [ x=\frac{-b}{2a} ]
Standard form is useful when you know the coefficients of the quadratic expression or when you are working with algebraic equations.
For example:
[ f(x)=2x^2-8x+6 ]
is a quadratic function in standard form. Here, (a=2), (b=-8), and (c=6).
Vertex Form of a Quadratic Function
The vertex form of a quadratic function is:
[ f(x)=a(x-h)^2+k ]
This form is especially helpful because it directly shows the vertex of the parabola:
[ (h,k) ]
The vertex is the highest or lowest point on the graph Easy to understand, harder to ignore..
- If (a>0), the vertex is the minimum point.
- If (a<0), the vertex is the maximum point.
For example:
[ f(x)=3(x-2)^2+5 ]
has vertex ((2,5)). Since (a=3), the parabola opens upward and is narrower than the parent function (f(x)=x^2).
Vertex form is one of the easiest ways to find a quadratic equation when you know the vertex and one other point on the graph Not complicated — just consistent..
Intercept Form of a Quadratic Function
The intercept form of a quadratic function is:
[ f(x)=a(x-p)(x-q) ]
In this form, (p) and (q) are the x-intercepts, also called roots or zeros of the function.
For example:
[ f(x)=2(x-3)(x+1) ]
has x-intercepts at:
[ x=3 \quad \text{and} \quad x=-1 ]
To find the value of (a), use another point on the graph and substitute its (x) and (y) values into the equation.
Intercept form is useful when you know the x-intercepts and one additional point.
Finding the Equation When You Know the Vertex and One Point
If you are given the vertex and one point, use vertex form It's one of those things that adds up..
Suppose the vertex is ((2,-3)), and the parabola passes through the point ((4,5)).
Start with vertex form:
[ f(x)=a(x-h)^2+k ]
Substitute the vertex ((2,-3)):
[ f(x)=a(x-2)^2-3 ]
Now use the point ((4,5)). Basically, when (x=4), (f(x)=5) Practical, not theoretical..
[ 5=a(4-2)^2-3 ]
Simplify:
[ 5=a(2)^2-3 ]
[ 5=4a-3 ]
Add 3 to both sides:
[ 8=4a ]
Divide by 4:
[ a=2 ]
So the equation is:
[ f(x)=2(x-2)^2-3 ]
This is the quadratic function in vertex form.
Finding the Equation When You Know Three Points
If you know three points on the parabola, you can use standard form and solve a system of equations.
Suppose the points are:
[ (0,3), \quad (1,2), \quad (2,3) ]
Start with standard form:
[ f(x)=ax^2+bx+c ]
Use the point ((0,3)):
[ 3=a(0)^2+b(0)+c ]
[ c=3 ]
Now the equation becomes:
[ f(x)=ax^2+bx+3 ]
Use the point ((1,2)):
[ 2=a(1)^2+b(1)+3 ]
[ 2=a+b+3 ]
[ a+b=-1 ]
Use the point ((2,3)):
[ 3=a(2)^2+b(2)+3 ]
[ 3=4a+2b+3 ]
[ 0=4a+2b ]
[ 2a+b=0 ]
Now solve the system:
[ a+b=-1 ]
[ 2a+b=0 ]
Subtract the first equation from the second:
[ a=1 ]
Substitute (a=1) into (a+b=-1):
[ 1+b=-1 ]
[ b=-2 ]
So the equation is:
[ f(x)=x^2-2x+3 ]
This method works for any three non-collinear points that lie on a quadratic graph.
Finding the Equation from a Graph
To find the equation of a quadratic function from a graph, look for important features:
- The vertex
- The y-intercept
- The x-intercepts
- Another point on the parabola
- Whether the parabola opens upward or downward
A strong strategy is to use the most useful form based on what the graph shows Not complicated — just consistent..
If the vertex is visible, use vertex form.
If the x-intercepts are visible, use intercept form Turns out it matters..
Using the Standard Form When the Y‑Intercept Is Known
The standard form of a quadratic function
[ f(x)=ax^{2}+bx+c ]
is especially handy when the y‑intercept ((0,c)) is evident on the graph.
Because the y‑intercept occurs when (x=0), the constant term (c) is simply the
(y)‑value at that point. Once (c) is fixed, any two additional points give a
system of two equations in the unknowns (a) and (b) Surprisingly effective..
Not obvious, but once you see it — you'll see it everywhere.
Example.
A parabola opens upward, has a y‑intercept at ((0,4)), and passes through
((1,6)) and ((3,20)). Find its equation.
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Write the standard form with the known intercept:
[ f(x)=ax^{2}+bx+4 ]
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Plug in ((1,6)):
[ 6=a(1)^{2}+b(1)+4 ;\Longrightarrow; a+b=-2 ]
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Plug in ((3,20)):
[ 20=a(3)^{2}+b(3)+4 ;\Longrightarrow; 9a+3b=16 ]
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Solve the linear system
[ \begin{cases} a+b=-2\[2pt] 9a+3b=16 \end{cases} ]
Multiply the first equation by 3: (3a+3b=-6).
Subtract from the second: ((9a+3b)-(3a+3b)=16-(-6)) → (6a=22) → (a=\dfrac{11}{3}).
Then (b=-2-\dfrac{11}{3}= -\dfrac{17}{3}) Small thing, real impact.. -
The quadratic is
[ f(x)=\frac{11}{3}x^{2}-\frac{17}{3}x+4. ]
Converting Between Forms
Knowing how to move from one representation to another gives you flexibility when solving problems.
| From → To | Key Steps |
|---|---|
| Intercept → Standard | Expand (a(x-p)(x-q)) to (ax^{2}+[-a(p+q)]x+apq). |
| Vertex → Standard | Expand (a(x-h)^{2}+k) to (ax^{2}+(-2ah)x+(ah^{2}+k)). |
| Standard → Vertex | Complete the square: (a[x^{2}+ \frac{b}{a}x] + c = a\bigl[(x+\frac{b}{2a})^{2}-\frac{b^{2}}{4a^{2}}\bigr] + c). |
| Standard → Intercept | Solve (ax^{2}+bx+c=0) for the roots (quadratic formula). |
Quick check: After converting, verify that the coefficients satisfy the original conditions (e.g., plug the known intercepts back into the expanded form).
A Full‑Graph Example
Suppose a parabola opens downward, its vertex is at ((-1,7)), and it crosses the x‑axis at (x=3) and (x=5). Find its equation.
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Identify the most convenient form.
The x‑intercepts are given, so start with intercept form:[ f(x)=a(x-3)(x-5). ]
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Use the vertex to solve for (a).
The vertex lies on the axis of symmetry, which is the midpoint of the intercepts:[ h = \frac{3+5}{2}=4. ]
On the flip side, the given vertex is ((-1,7)), not ((4,?)). That said, this tells us that the parabola is reflected about its axis; the vertex is not the midpoint of the intercepts because the parabola is not symmetric about the line (x=4). Therefore we must use the vertex directly.
Plug the vertex ((-1,7)) into the intercept form:
[ 7 = a(-1-3)(-1-5) = a(-4)(-6) = 24a ;\Longrightarrow; a=\frac{7}{24}. ]
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Write the final equation.
[ f(x)=\frac{7}{24}(x-3)(x-5). ]
