# Quadratic functions define quadratic function . A quadratic function by bfk20410

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```									Oct 22 Quadratic Functions
Homework: 4.2 #5, 17, 19, 27, 33, 39, 43, 45, 47, 53, 55

Quadratic functions
define quadratic function. A quadratic function is a function defined by a
quadratic polynomial, f ( x ) = a2 x 2 + a1 x + a0 where a2 , a1, a0 constants
with a2 ! 0 or (more commonly) f ( x ) = ax 2 + bx + c where a, b, c
constants with a ! 0 . E.g., y = !2x 2 + 3x !1.
define quadratic-like functions. Some functions have a similar form to quadratic
functions, e.g., f ( x ) = !2x + 3 x !1, g( x ) = !2x 4 + 3x 2 !1, and
h ( x ) = !2e 2 x + 3e x !1. Each of these functions can be transformed into a
quadratic function by using an appropriate substitution, e.g., in g we can
substitute y = x 2 to get g( y ) = !2y 2 + 3y !1. We can now find the zeros
of g( x ) = !2x 4 + 3x 2 !1 by first finding the zeros of g( y ) = !2y 2 + 3y !1,
1                                                  1
and we find y = 1, y = . We back substitute for y to get x 2 = 1, x 2 = .
2                                                  2
2
Solving for x, we get x = ±1, x = ±        .
2
graphs of quadratic functions, parabola, vertex, axis of symmetry. Graphs of
quadratic functions are curves called parabolas. Every parabola has two
important features: an axis of symmetry and a vertex. The vertex is either
a global maximum or minimum point and the axis of symmetry is a
vertical line that passes through the vertex. The graph of every quadratic
function also has a y-intercept and either zero, one or two x-intercepts.
Find the equation of the parabola with a vertical axis of symmetry that
passes through the points (1,0), (2,9), and (!3,4 ) .
translations of y = x 2 , vertex quadratic form. Draw graph of y = x 2 . We have
sketched the graph of y = !x 2 + 2x + 2 previously by converting the
equation to y = !( x !1) + 3, where the vertex is (1,3) . This suggests
2

another quadratic form (called the vertex form), i.e., y = a( x ! h) 2 + k ,
where (h,k) is the vertex and x = h is the equation of the axis of
symmetry. If a > 0 , the parabola opens upward; if a < 0 , it opens
downward. The vertex form is useful for finding the vertex and the axis of
symmetry easily.
vertex form of quadratic functions. We can find general formulas for calculating
the vertex and axis of symmetry for f (x) = ax 2 + bx + c by completing its
square, i.e., f (x) = ax 2 + bx + c =
b                 b     b              b          b         b2
a(x 2 +     x) + c = a(x 2 + x + [ ]2 ) + c ! a[ ]2 = a(x + ) 2 + c !    .
a                 a     2a            2a         2a         4a
b      b2
This is now in vertex form and the vertex is the point (!      , c ! ) and
2a      4a
b
the axis of symmetry is x = !      . Find the equation whose
2a
graph is a parabola with vertex ( 4,!3) , a vertical axis of symmetry, and
passes through the point (5,!6) .
factored form of quadratic functions, discriminant. The quadratic formula,
!b ± b 2 ! 4ac
x=                  , calculates zeroes of the function f ( x) = ax 2 + bx + c .
2a
These zeroes (or roots of the equation ax 2 + bx + c = 0 ) may be real or
complex. Let r1 and r2 represent the roots of the equation, then we can
rewrite f(x) as f ( x) = a( x ! r1 )( x ! r2 ) . Early we stated that the roots,
r1 and r2 , could be two distinct real roots, one repeated real root, or two
complex roots. The discriminant, b 2 ! 4ac , will determine what type of
roots we have. If b 2 ! 4ac > 0 , then we have two distinct real roots. If
b 2 ! 4ac = 0 , one repeated real root. If b 2 ! 4ac < 0 , two complex roots.
Why? Notice that complex roots always occur in conjugate pairs, i.e.,

b       b 2 ! 4ac
x =! ±                  i . Find the quadratic function whose graph has
2a          2a
zeros at x = !5 and x = 3 and a y-intercept at (0, 30). Using the general
form, f ( x) = ax 2 + bx + c substitute the points (!5,0), ( 3,0), and (0,30) for
x and y. We obtain three functions:
25a ! 5b + c = 0
9a + 3b + c = 0
c = 30
Solving for A, B, and C, A = -2, B = -4, and C = 30. So, y = !2x 2 ! 4 x + 30 .
Using the vertex form. We know that h lies on the axis of symmetry,
which is the midpoint of the two zeros, so h = !1 . So, y = a( x +1) 2 + k .
Again, after substituting our three points, we obtain 3 functions:
16a + k = 0
16a + k = 0
a + k = 30
Solving for a and k, we obtain a = !2 and k = 32 . So, y = !2( x +1) 2 + 32 .
Using the factored form, f ( x) = a( x ! r1 )( x ! r2 ) . We know r1 and r2 .
Therefore, y = a( x + 5)( x ! 3) . Substituting the point (0, 30), we obtain
!15a = 30 , so a = -2 and y = !2( x + 5)( x ! 3) . Show that
!2( x + 5)( x ! 3) = !2x 2 ! 4 x + 30 = !2( x +1) 2 + 32 .
r1 + r2
compute vertex and axis of symmetry. The axis of symmetry is x =                 (the
2
average of the two roots). The y-value of the vertex can be found by
r1 + r2                                            a
substituting x =        into f ( x) and calculating its value, ! (r1 ! r2 ) 2 .
2                                               4
r1 + r2 a
Therefore, the vertex is (        ,! (r1 ! r2 ) 2 ) . Compute the vertex and axis
2      4
of symmetry of y = !2( x + 5)( x ! 3) .
compute intercepts. To find the x-intercepts of the graph of any function f (if they
exist), we solve the equation f ( x ) = 0 . To find the y-intercept, we
calculate the value f (0) . Find the intercepts of y = !2x 2 ! 4 x + 30 .
application. Find the equation of the tangent lines to the circle x 2 + y 2 = 10 from
the point (1,!7) .

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