Exercise 12 Page 604

To draw the graph of the given quadratic function written in standard form, we must start by identifying the values of a, b, and c. y=3x^2+x-5 ⇔ y=3x^2+1x+(- 5) We can see that a=3, b=1, and c=- 5. Now, we will follow three steps to graph the function.

1. Find the axis of symmetry.
2. Calculate the vertex.
3. Graph the function.

   Finding the Axis of Symmetry

   The axis of symmetry is a vertical line with equation x=- b2a. Since we already know the values of a and b, we can substitute them into the formula.

   x=- b/2a

   SubstituteII

   a= 3, b= 1

   x=- 1/2(3)

   Multiply

   Multiply

   x=-1/6

   The axis of symmetry of the parabola is the vertical line with equation x=- 16.

   Calculating the Vertex

   To calculate the vertex, we need to think of y as a function of x, y=f(x). We can write the expression for the vertex by stating the x- and y-coordinates in terms of a and b. Vertex: ( - b/2a, f( - b/2a ) ) Note that the formula for the x-coordinate is the same as the formula for the axis of symmetry, which is x=- 16. Thus, the x-coordinate of the vertex is also - 16. To find the y-coordinate, we need to substitute - 16 for x in the given equation.

   y=3x^2+x-5

   Substitute

   x= -1/6

   y=3( -1/6)^2+( -1/6)-5

   ▼

   Simplify right-hand side

   PowQuot

   (a/b)^m=a^m/b^m

   y=3*1/36+ ( - 1/6 )-5

   AddNeg

   a+(- b)=a-b

   y=3*1/36- 1/6 -5

   MoveLeftFacToNumOne

   a* 1/b= a/b

   y=3/36-1/6-5

   ReduceFrac

   a/b=.a /3./.b /3.

   y=1/12-1/6-5

   ExpandFrac

   a/b=a * 2/b * 2

   y=1/12-2/12-5

   NumberToFrac

   a = 12* a/12

   y=1/12-2/12-60/12

   SubFrac

   Subtract fractions

   y=-61/12

   We found the y-coordinate, and now we know that the vertex is (- 16,- 6112).

   Graphing the function

   To graph the function we will make a table of values.

   x	3x^2+x-5	y=3x^2+x-5
   -2	3( -2)^2+( -2)-5	5
   -1	3( -1)^2+( -1)-5	-3
   1	3( 1)^2+ 1-5	-1
   2	3( 2)^2+ 2-5	9

   Now, let's plot the obtained points, then we can connect them with a parabola.