Exercise 1 Page 148

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=x^2-2x-8 ⇔ y=1x^2+-2x+(- 8) We can see that a=1, b=-2, and c=- 8. Now, we will follow four steps to graph the function.

1. Find the axis of symmetry.
2. Calculate the vertex.
3. Identify the y-intercept and its reflection across the axis of symmetry.
4. Connect the points with a parabola.

   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= 1, b= -2

   x=- -2/2(1)

   ▼

   Simplify right-hand side

   Multiply

   Multiply

   x=- (-2/2 )

   MoveNegNumToFrac

   Put minus sign in front of fraction

   x=- ( - 2/2 )

   NegNeg

   - (- a)=a

   x=2/2

   CalcQuot

   Calculate quotient

   x=1

   The axis of symmetry of the parabola is the vertical line with equation x=1.

   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=1. Thus, the x-coordinate of the vertex is also 1. To find the y-coordinate, we need to substitute 1 for x in the given equation.

   y=x^2-2x-8

   Substitute

   x= 1

   y=( 1)^2-2( 1)-8

   ▼

   Simplify right-hand side

   CalcPow

   Calculate power

   y=1-2(1)-8

   Multiply

   Multiply

   y=1-2-8

   AddSubTerms

   Add and subtract terms

   y=-9

   We found the y-coordinate, and now we know that the vertex is (1,-9).

   Identifying the y-intercept and its Reflection

   The y-intercept of the graph of a quadratic function written in standard form is given by the value of c. Thus, the point where our graph intercepts the y-axis is (0,- 8). Let's plot this point and its reflection across the axis of symmetry.

   

   Connecting the Points

   We can now draw the graph of the function. Since a=1, which is positive, the parabola will open upwards. Let's connect the three points with a smooth curve.

   

   Now that we have the graph of the function, we can solve x^2-2x-8=0. Notice that this is where our graph intersects the line y=0. From our graph, we can see that this happens at x=-2 and x=4.