Transcript/Graphics for the Most Complicated Problem in Intermediate Algebra
Transcript:
The function, y = –2x2 – 8x – 6, is a second-degree polynomial called a
quadratic function. The graph of a quadratic function is a parabola. It has
a vertex, a highest or lowest point. It also has one y-intercept and possibly one
or two x-intercepts.
graphic:
y = –2x2 – 8x – 6
Transcript:
If the equation of a parabola is given in quadratic form, y  ax 2  bx  c ,
then the x-value of the vertex is given by the formula  b / 2a where b is the
coefficient to the linear term of the polynomial and a is the leading coefficient.
graphics:
quadratic form: y  ax 2  bx  c
formula for x-value of vertex of parabola in quadratic form:  b / 2a
Transcript:
In this case, b equals negative eight, and a equals negative two. So, the
formula  b / 2a yields –2.
graphics:
y=
–2x – 8x – 6
y = –2x – 8x – 6
2
2
 b  ( 8)
8


 2
2a 2 2  4
Remember that the vertex is a point, the highest or lowest point of the
parabola. Consequently, it has a x-value and a y-value. To find the
y-value of the vertex, substitute the x-value into the parabola's equation.
When we substitute negative two in for x, the given equation yields two
for the y-value of the vertex. Thus the vertex is the ordered pair, (–2,2).
x  value of vertex 
Transcript:
graphics:
Find y-value of vertex by substituting in x-value of vertex into equation:
y  2( 2) 2  8( 2)  6
y  24  16  6
y  8  16  6
y2
(–2,2)
Transcript:
Once we have the vertex, we should find the y-intercept. The y-intercept
is the value of the function when x equals zero. We can therefore find the
y-intercept by substituting zero for x. When the parabola is written in
quadratic form, y  ax 2  bx  c , the y-intercept will always be the
ordered pair (0,c). In the given equation y = –2x2 – 8x – 6, c equals
negative six, so the y-intercept is the point (0,-6).
graphics:
y  ax 2  bx  c
(0,c)
y = –2x2 – 8x
–6
y-intercept: (0,–6)
Transcript:
Graphing the vertex and the y-intercept and then sketching the parabola
reveals that the graph will have two x-intercepts. We see this for two
reasons. First, whenever a parabola in quadratic form, y  ax 2  bx  c ,
has a negative leading coefficient (in other words, a is negative), the
parabola opens down. Conversely, if the leading coefficient is positive,
the parabola opens up. In this case, the leading coefficient equals negative
two, so the parabola opens down from the vertex (-2,2); therefore, it must
have x-intercepts. Second, if the vertex and y-intercept are on opposite
sides of the x-axis, then there must be x-intercepts since the graph will
cross the x-axis from the vertex to the y-intercept. In this case, since the
vertex is above the x-axis and the y-intercept is below the x-axis, there
must x-intercepts.
graphics:
quadratic form: y = ax2 + bx + c
leading coefficient: y =
a
x2 + bx + c
negative leading coefficient, parabola opens down:
positive leading coefficient, parabola opens up:
vertex: (-2,2)
·
·
y-intercept: (0,-6)
Note to Joe: For the above graphic the vertex should appear, then the y-intercept, then the
curve. I had a hard time getting the curve to go through the points. You can tell they are off a
little.
Transcript:
Since the graph has two x-intercepts, they should be labeled for the sketch
to be considered complete. Thus, we must find the x-intercepts. These
intercepts occur where y equals zero. Accordingly, we can substitute zero
for y in the equation and solve the resulting equation. The x-values that solve a
quadratic equation set equal to zero are the intercepts, often called roots or zeros.
graphics:
y = –2x2 – 8x – 6
0 = –2x2 – 8x – 6
–2x2 – 8x – 6 = 0
Transcript:
To solve a quadratic equation, we can use completing the square or the quadratic
formula. If the solutions are rational, then the quadratic equation can be solved by
factoring. In this case, we will first multiply every term in the equation by
negative one, changing the equation from –2x2 – 8x – 6 = 0 to 2x2 + 8x + 6 = 0.
Next, we can factor out the greatest common factor or simply divide every term in
the equation by the greatest common factor, changing the equation from
2x2 + 8x + 6 = 0 to x2 + 4x + 3 = 0. The trinomial x2 + 4x + 3 factors to the
product of the quantities (x + 1) and (x + 3), so we can rewrite the equation from
to x2 + 4x + 3 = 0 to (x + 1)(x + 3) = 0. We can now set each factor equal to zero,
giving us two linear equations. First, x + 1 = 0. Second, x + 3 = 0. Solving these
two linear equations gives us the roots of the quadratic: x = –1 and x = –3. Thus,
the x-intercepts are (–1,0) and (–3,0).
graphics:
–2x2 – 8x – 6 = 0
 1   2 x 2  8 x  6  0  1
2 x 2  8x  6  0
2 x 2 8x 6 0

 
2
2 2 2
x2  4x  3  0
x  1( x  3)  0
x+1=0
x+3=0
x = –1
x = –3
(–1,0) and (–3,0)
Transcript:
We can now complete the sketch of the graph by labeling x-intercepts. Once the
with the vertex and all intercepts labeled, the graph is complete.
graphics:
(-2,2)
(-1,0)
(-3,0)
(0,-6)
Graphics without Transcripts for the Most Complicated Problem in Intermediate Algebra
graphic:
y = –2x2 – 8x – 6
quadratic form: y  ax 2  bx  c
formula for x-value of vertex of parabola in quadratic form:  b / 2a
–2x – 8x – 6
y = –2x – 8x – 6
2
y=
2
 b  ( 8)
8


 2
2a 2 2  4
Find y-value of vertex by substituting in x-value of vertex into equation:
y  2( 2) 2  8( 2)  6
y  24  16  6
y  8  16  6
y2
(–2,2)
y  ax 2  bx  c
(0,c)
x  value of vertex 
–6
y = –2x2 – 8x
y-intercept: (0,–6)
quadratic form: y = ax2 + bx + c
leading coefficient: y =
a
x2 + bx + c
negative leading coefficient, parabola opens down:
positive leading coefficient, parabola opens up:
vertex: (-2,2)
·
·
y-intercept: (0,-6)
y = –2x2 – 8x – 6
0 = –2x2 – 8x – 6
–2x2 – 8x – 6 = 0
–2x2 – 8x – 6 = 0
 1   2 x 2  8 x  6  0  1
2 x 2  8x  6  0
2 x 2 8x 6 0

 
2
2 2 2
x2  4x  3  0
x  1( x  3)  0
x+1=0
x+3=0
x = –1
x = –3
(–1,0) and (–3,0)
(-2,2)
(-1,0)
(-3,0)
(0,-6)