# Convert quadratic functions from one form to another ```CONVERT QUADRATIC FUNCTIONS FROM ONE FORM TO ANOTHER
(Standard Form &lt;==&gt; Intercept Form &lt;==&gt; Vertex Form)
(By Nghi H Nguyen – Dec 08, 2014)
1. THE QUADRATIC FUNCTION IN INTERCEPT FORM
The graph of the quadratic function in standard form, f(x) = ax&sup2; + bx + c, is a parabola that may
intercept the x-axis at 2 points, unique point, or no point at all. This means a quadratic equation
f(x) = ax&sup2; + bx + c = 0 may have 2 real roots, one double real root, or no real roots at all
(complex roots).
We can write f(x) in the form: y = a(x&sup2; + bx/a + c/a) (1).
Recall the development of the quadratic formula:
x&sup2; + bx/a + (b&sup2;/4a&sup2; - b&sup2;/4a&sup2;) + c/a = 0
(x + b/2a)&sup2; - (b&sup2; - 4ac)/4a&sup2; = 0
(x + b/2a)&sup2; - d&sup2;/4a&sup2; = 0.
(Call d&sup2; = b&sup2; - 4ac)
(x + b/2a + d/2a)(x + b/2a – d/2a) = 0 (2)
Replace this expression (2) into the equation (1), we get the quadratic function f(x) written in
intercept form:
f(x) = a(x – x1)(x – x2)
(x1 and x2 are the 2 x-intercepts, or roots of f(x) = 0)
f(x) = a(x + b/2a + d/2a)(x + b/2a – d/2a) (3)
2. THE QUADRATIC FORMULA IN INTERCEPT FORM
From the equation (3), we deduct the formula:
x = -b/2a &plusmn; d/2a (4)
In this formula, x being the 2 roots of the quadratic equation f(x) = 0:
-
The quantity (–b/2a) represents the x-coordinate of the parabola axis.
The 2 quantities (d/2a) and (-d/2a) represent the 2 distances AB and AC from the
parabola axis to the two x-intercepts (real roots) of the parabola.
The quantity d can be zero, a number (real or radical), or imaginary number.
If d = 0: there is a double root at x = -b/2a.
If d is a number (real or radical), there are two x-intercepts, meaning two real roots.
-
If d is imaginary, there are no real roots. The parabola doesn’t intercept the x-axis.
-
The unknown quantity d can be computed from the constants a, b, c by this relation (5):
d&sup2; = b&sup2; – 4ac
(5)
We can easily obtain this relation by writing that the product (x1*x2) of the 2 real roots is equal
to (c/a).
(-b/2a + d/2a)(-b/2a – d/2a) = c/a
b&sup2; – d&sup2; = 4ac  d&sup2; = b&sup2; – 4ac
If d &sup2; = 0, there is double root at x = -b/2a
If d&sup2; &gt; 0, there are 2 real roots.
If d&sup2; &lt; 0, there are no real roots, there are complex roots.
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To solve a quadratic equation, first find the quantity d by the relation (5), then find the 2 real
roots by the formula (4).
REMARK. This new quadratic formula in intercept (graphic) form is simpler and easier to
remember than the classic formula since students can relate it to the x-intercepts of the
parabola graph. In addition, the two quantities (d/2a) and (-d/2a) make more sense about
distance than the classical quantity: √(b^2 – 4ac).
3. CONVERT QUADRATIC EQUATIONS FROM STANDARD FORM TO VERTEX FORM
Problem. Given f(x) = ax&sup2; + bx + c.
Find x1 and x2 to convert f(x) to intercept form: f(x) = a*(x – x1)(x – x2)
Solution. To convert a function, from standard form f(x) = ax&sup2; + bx + c, to intercept form, we
use the general intercept form expression (3).
f(x) = a*(x + b/2a + d/2a)(x + b/2a – d/2a), with d&sup2; = b&sup2; - 4ac.
a. If d is real, whole number (∓3, ∓14, ∓76…), or fraction (∓3/5, ∓23/47…), there are 2 xintercepts. f(x) can be factored and f(x) can be converted to intercept form.
Examples: f(x) = (x – 7)(x + 8); y = (x + 3)(x + 23);
f(x)= (3x – 5)(7x – 13); y = (5x – 7)(6x – 5)
b. If d is radical (∓√5, ∓√13, ∓√71…), there are 2 x-intercepts. f(x) can’t be factored.
However f(x) can be converted to intercept form. See Example 10 and 11.
Examples: f(x) = (x – 1/2 + √5/2)(x – 1/2 - √5/2); f(x) = (x - 2/3 - √3/3)(x – 2/3 + √3/3)
c. If d is imaginary (∓3i, ∓17i…), there are no x-intercepts, and there is no intercept
form.
Example 1. Convert y = 5x&sup2; + 6x – 8 into intercept form
Solution. First find d&sup2; = 36 + 160 = 196  d = 14 and d = - 14. Substitute the values of d, a, b, c
into (3), we get y in intercept form.
y = 5*(x + 6/10 + 14/10)(x + 6/10 – 14/10)
y = 5*(x – 4/5)(x + 2) = (5x – 4)(x + 2).
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Example 2. Convert y = 7x^2 + 18x – 25 to intercept form
Solution. First find d&sup2; = 324 + 700 = 1024  d = 32 and d = -32.
y = 7*(x + 18/14 - 32/14)(x + 18/14 + 32/14)
Intercept form: y = 7*(x - 1)(x + 25/7) = (x – 1)(7x + 25)
Example 3. Convert y = 16x&sup2; + 34x – 15 to intercept form.
Solution. First find d&sup2; = b&sup2; – 4ac = 1156 + 960 = 2116  d = 46 and d = -46
Since d is a whole number, we can either factor the equation, or use the formula.
Factoring gives the intercept form: y = (8x – 3)(2x + 5)
Example 4. Convert y = 9x&sup2; + 4x – 1 to intercept form
Solution. d&sup2; = 16 + 36 = 52 = 4*13  d = 2√13 and d = -2√13.
Replace values of d, a, b, c into the general expression (3).
Intercept form: y = 9[x + (2 + √13)/9)(x + (2 - √13)/9).
y = 1/9(9x + 2 + √13)(9x + 2 - √13)
Back check. We may check the answer by multiplying the 2 factors:
y = 1/9[(9x + 2)&sup2; - 13] = 1/9*(81x&sup2; + 36x + 4 – 13) = 9x&sup2; + 4x - 1
Example 5. Convert y = 9x&sup2; - 12x + 1 to intercept form.
Solution. d&sup2; = 144 – 36 = 108 = 36*3  d = 6√3 and d = -6√3.
Intercept form: y = 9(x - 2/3 + √3/3)(x - 2/3 - √3/3) = (3x – 2 + √3)(3x – 2 - √3)
Back check. We may check the answer by multiplying the 2 factors:
y = (3x – 2)&sup2; - 3 = 9x&sup2; - 12x + 4 – 3 = 9x&sup2; - 12x + 1
4. CONVERT QUADRATIC FUNCTIONS FROM STANDARD FORM TO VERTEX FORM
We can convert a quadratic function from standard form, y = ax&sup2; + bx + c, to the general vertex
form: y = a(x + p)&sup2; + q. We don’t need to factor the quadratic equation because factoring is only
a special case of finding the 2 real roots. The below method is generally better.
Problem. Given a quadratic function in standard form: f(x) = ax&sup2; + bx + c
Find p and q to convert f(x) to the Vertex Form: f(x) = a*(x + p)&sup2; + q
Recall the formula (4) of the quadratic formula in intercept form:
x = -b/2a ∓ d/2a. (4)
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Solution. The quantity (-b/2a) represents the x-coordinate of the vertex. At the vertex, d = 0
since the horizontal distances of the vertex to the 2 related x-intercepts are zero. The ycoordinate of the vertex is f(-b/2a). Consequently, the general vertex form of a quadratic
function can be written as:
f(x) = a(x + b/2a)&sup2; + f(-b/2a) (6)
Example 1. Convert y = 5x&sup2; + 6x – 8 to vertex form.
Solution. We have: a = 5; and (-b/2a) = -6/10 = -3/5;
f(-3/5) = ax&sup2; + bx + c = 5(9/25) – 6(3/5) – 8 = -49/5
The vertex form  y = 5(x + 3/5)&sup2; - 49/5
Example 2. Convert y = 7x&sup2; + 18x – 25 to vertex form.
Solution. We have: a = 7; and (-b/2a) = -18/14 = -9/7
f(-9/7) = 7(81/49) – 18(9/7) – 25 = 81/7 – 162/7 – 175/7 = 256/7
The vertex form  y = 7(x + 9/7)&sup2; - 256/7.
Example 3. Convert y = 2x&sup2; + 9x + 10, to vertex form.
Solution. (-b/2a) = -9/4
f(-9/4) = 2(81/16) – 9(9//4) + 10 = 81/8 – 162/8 + 80/8 = -1/8.
Vertex form  y = 2(x + 9/4)&sup2; - 1/8.
NOTE. There is another method to convert y, from standard form to vertex form, by completing
the squares. Both methods, using general expression or completing the squares, start from the
same origin: the development of the quadratic formula.
Example 4. Convert y = 5x&sup2; + 6x – 8 to vertex form.
Solution. Write y in the form:
y = 5*[x&sup2; + 6x/5 + (9/25 – 9/25) – 8/5] = 5(x + 3/5)&sup2; - 5*(49/25)
y = 5*(x + 3/5)&sup2; - 49/5.
Example 5. Convert y = 5x&sup2; - 8x + 2 to vertex form.
Solution. y = 5*(x&sup2; - 8x/5 + 2/5) = 5(x – 4/5)&sup2; – 16/5 + 2
Vertex form; y = 5*(x – 4/5)&sup2; - 6/5
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5. CONVERT QUADRATIC FUNCTIONS FROM INTERCEPT FORM TO STANDARD FORM.
Problem. Given a quadratic function in intercept form: f(x) = a*(x – x1)(x – x2) (1)
Find a, b, c to convert f(x) to standard form: f(x) = ax&sup2; + bx + c. (2)
Reminder:
a.- f(x) = ax&sup2; + bx + c can be factored If x1 and x2 are real (whole number or fraction).
Examples: x = -4; x = 23; x = 4/7; x = -7/13
There are two x-intercepts, and f(x) can be in intercept form
b.- f(x) can’t be factored If x1 and x2 are radical [(2 + √3), (2 - √3)…].
Examples: x1 = 1/2 + √3/2; x2 = 1/2 - √3/2; x1 = 3 - √5/, x2 = 3 + √5.
However, there are two x-intercepts, and f(x) can be in intercept form. See Examples 5 and 6.
Recall:
Sum of the roots (x1 + x2) = -b/a (3), and Product of the roots (x1*x2) = c/a. (4)
Solution. a is given by form (1), directly or indirectly.
b is obtained from relation (3). The two x-intercepts x1 and x2 are given in the form (1)
c can be found from relation (4).
Example 1. Convert f(x) = (x – 3)(x + 5) to standard form.
Solution. We have: a = 1; x1 = 3, and x2 = -5  (x1 + x2) = -2 = -b  b = 2.
x1*x2 = 3*(-5) = -15 = c  c = -15.
Answer. Standard form: x&sup2; + 2x – 15.
Example 2. Convert f(x) = (5x – 4)(x + 2) to standard form.
Solution. Write f(x) in the form: f(x) = 5*(x – 4/5)(x + 2).
We get a = 5; x1 = 4/5; and x2 = -2;
Relation (3) gives: -b/a = -b/5 = (x1 + x2) = (4/5 – 2) = -6/5  b = 6.
Relation (4) gives: c/a = c/5 = x1*x2 = (4/5)*(-2) = -8/5  c = -8.
Answer. Standard form: f(x) = 5x&sup2; + 6x – 8.
Example 3. Convert y = (x + 1)(7x – 25) to standard form.
Solution. y = 7(x + 1)(x -25/7)
We have: a = 7; x1 = -1; and x2 = 25/7.
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(x1 + x2) = -1 + 25/7 = -b/7  b = -18.
(x1)*(x2) = (-1)(25/7) = -25/7 = c/7  c = -25
Answer. Standard Form: y = 7x&sup2; - 18x – 25.
Example 4. Convert y = (8x – 3)(2x + 5) to standard form.
Solution. We have: x1 = 3/8; and x2 = -5/2; a = 2*(8) = 16; and c = 3*(-5) = -15
(x1 + x2) = 5/2 – 3/8 = 20/8 – 3/8 = 17/ 8 = -b/a = -b/16  b = -34.
Answer. Standard form: y = 16x^2 – 34x - 15.
Example 5. Convert f(x) = 3*(x + 2/3 - √5/3)(x + 2/3 + √5/3) to standard form.
Solution. We have a = 3; x1 = -2/3 + √5/3; and x2 = -2/3 - √5/3
(x1 + x2) = -2/3 – 2/3 = -4/3 = -b/a = -b/3  b = 4.
x1*x2 = (2/3)&sup2; – (√5/3)&sup2; = 4/9 – 5/9 = -1/9 = c/3  c = -3/9 = -1/3
Answer. Standard form f(x) = 3x&sup2; + 4x – 1/3.
Example 6. Convert f(x) = (x – 2/3 + √3/3)(x – 2/3 - √3/3) to standard form.
Solution. We have: a = 3*(3) = 9;
x1*x2 = (2/3)&sup2; – 3/9) = 4/9 – 3/9 = 1/9 = c/a  c = 9/9 = 1
(x1 + x2) = 4/3 = -b/a = -b/9 =  b = -12
Answer. Standard form: y = 9x&sup2; - 12x + 1.
6. CONVERT QUADRATIC FUNCTIONS FROM VERTEX FORM TO STANDARD FORM
Problem: Given Vertex Form: f(x) = a*(x + b/2a)&sup2; + f(-b/2a). (1)
Find a, b, c to convert f(x) to Standard Form: f(x) = ax&sup2; + bx + c. (2)
Solution. First, a is shown in form (1).
The quantity (-b/2a) that shows the x-coordinate of the parabola-axis is given in the
parentheses. This quantity gives the value of b. The constant c can be obtained from this
standard form: f(x) = ax&sup2; + bx + c.
f(-b/2a) = a*(-b/2a)&sup2; + b(-b/2a) + c. (3), where x is replaced by (-b/2a)
Example 1. Convert f(x) = (x + 1)&sup2; - 16 to standard form.
Solution. a = 1; and (-b/2a) = -1  b = 2
f(-1) = -16 = (-1)&sup2; + 2(-1) + c  c = -16 + 1 = -15.
Answer: Standard form: f(x) = x&sup2; + 2x – 15.
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Example 2. Convert f(x) = 5(x + 3/5)&sup2; – 49/5 to standard form.
Solution. We get a = 5; and (-b/2a) = -b/10 = -3/5  b = 6.
f(-3/5) = -49/5 = 5(-3/5)&sup2; + 6(-3/5) + c  c = -49/5 – 9/5 + 18/5 = -40/5 = -8.
Answer. Standard form: f(x) = 5x&sup2; + 6x – 8.
Example 3. Convert f(x) = 7(x + 9/7)&sup2; – 256/7 to standard form.
Solution. We get: a = 7; and (-b/2a) = -b/14 = -9/7  b = 18.
f(-9/7) = -256/7 = 7*(-9/7)&sup2; + 18(-9/7) + c.
c = -256/7 - 81/7 + 162/7 = -175/7 = -25
Answer. Standard form: f(x) = 7x&sup2; + 18x – 25.
NOTE. There is another method to convert the vertex form, y = a*(x + b/2a)&sup2; + f(-b/2a) to
standard form, y = ax^2 + bx + c. This method expands the expression in the parentheses.
Example 4. Convert y = 5*(x + 3/5)&sup2; - 49/5.
Solution. Expand y = 5*( x&sup2; + 6x/5 + 9/25) = (5x&sup2; + 6x + 9/5) – 49/5 = 5x^2 + 6x – 8.
Example 5. Convert y = 7*(x + 9/7)&sup2; - 256/7 to standard form.
Solution. Develop the expression in the parentheses:
y = 7*(x&sup2; + 18x/7 + 81/49) – 256/7 = 7x&sup2; + 18x + 81/7 – 256/7
Answer. Standard form: y = 7x&sup2; + 18x – 25
7. CONVERT QUADRATIC FUNCTIONS FROM INTERCEPT FORM TO VERTEX FORM.
Problem. Given a quadratic function in intercept form: f(x) = a*(x – x1)(x – x2) (1)
Find (-b/2a) to convert f(x) to vertex form: f(x) = a(x + b/2a)&sup2; + f(-b/2a) (2)
Solution. We get value of a from Relation (1), directly or indirectly. We get b from relation: (x1
+ x2) = -b/a (3). Find f(-b/2a) by Relation (1) to complete Form (2).
Example 1. Convert f(x) = (5x – 4)(x + 2) to vertex form.
Solution. We have a = 5. Write f(x) in the form:
f(x) = 5*(x – 4/5)(x + 2). We get: x1 = 4/5; and x2 = -2.
(x1 + x2) = 4/5 – 10/5 = -6/5 = -b/a  b = 6  (-b/2a) = -6/10 = -3/5
Replace x by (-3/5) into Relation (1).
Find f(-b/2a) = f(-3/5) = 5*(-3/5 – 4/5)(-3/5 + 2) = 5*(-7/5)(7/5) = -49/5.
Answer. Vertex form: f(x) = 5(x + 3/5)&sup2; – 49/5.
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Example 2. Convert f(x) = (7x + 25)(x - 1) to vertex form
Solution. a = 7; x1 = -25/7; and x2 = 1.
(x1 + x2) = -25/7 + 7/7 = -18/7 = -b/a = -b/7  b = 18  -b/2a = 18/14 = -9/7
f(-b/2a) = f(-9/7) = (-9 + 25)(-9/7 - 1) = 16*(- 16/7) = -256/7
Answer. Vertex form: f(x) = 7*(x + 9/7)&sup2; - 256/7.
Example 3. Convert f(x) = (x + 2)(2x + 5) to vertex form.
Solution. f(x) = 2*(x + 2)(x + 5/2)  a = 2; x1 = -2; and x2 = -5/2.
(x1 + x2) = -2 – 5/2 = -9/2 = -b/a = -b/2  b = 9  -b/2a = -9/4.
f(-9/4) = (-9/4 + 8/4)(-18/4 + 20/4) = (-1/4)(2/4) = -2/16 = -1/8.
Answer. Vertex form: f(x) = 2*(x + 9/4)&sup2; - 1/8.
NOTE. When the intercept form f(x) is radical, we can also convert f(x) to vertex form.
Example 4. Convert y = (3x – 2 + √3)(3x – 2 - √3) to vertex form.
Solution. We have: a = 9; x1 = 2/3 + √3/3; and x2 = 2/3 - √3/3.
x-coordinate of parabola axis, or vertex: (x1 + x2)/2 = 4/6 = 2/3
y-coordinate of vertex: f(2/3) = (6/3 – 2 + √3)(6/3 – 2 - √3) = -3
Answer. Vertex form: 9(x – 2/3)&sup2; - 3.
Example 5. Convert y = 3*(x + 2/3 + √5/3)(x + 2/3 - √5/3) to vertex form.
Solution. a = 3; and x1 = -2/3 + √5/3; and x2 = -2/3 - √5/3
x-coordinate of vertex: (x1 + x2)/2 = -2/3
y-coordinate of vertex: f(-2/3) = 3*(√5/3)(-√5/3) = 3*(-5/9) = -5/3
Answer. Vertex form y = 3(x + 2/3)&sup2; - 5/3
[This article was written by Nghi H. Nguyen, author of the new Transforming Method for solving