Exercise 122 Page 49

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+(-2)x+(-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

   MultByOne

   a * 1=a

   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

   MultByOne

   a * 1=a

   y=1-2-8

   SubTerms

   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.

   

   The domain of quadratic functions is all real numbers. We can see above that the minimum point of the curve is reached at the vertex. Thus, the range is all real numbers greater than or equal to -9. Domain:& all real numbers Range:& y ≥ -9 By looking at the graph, we can state approximated values for the x-intercepts. We can see that the parabola intercepts the x-axis at -2 and 4, approximately.