Exercise 1 Page 94

We have a quadratic equation, written in standard form. To draw the graph of the related quadratic function we must start by identifying the values of a, b, and c. y=x^2-8x+12 ⇔ y=1x^2+(- 8)x+12 We can see that a=1, b=- 8, and c=12. 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= - 8

   x=- - 8/2(1)

   ▼

   Simplify right-hand side

   IdPropMult

   Identity Property of Multiplication

   x=- - 8/2

   RemoveNegFracAndNum

   - - a/b= a/b

   x=8/2

   CalcQuot

   Calculate quotient

   x=4

   The axis of symmetry of the parabola is the vertical line with equation x=4.

   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=4. Thus, the x-coordinate of the vertex is also 4. To find the y-coordinate, we need to substitute 4 for x in the given equation.

   y=x^2-8x+12

   Substitute

   x= 4

   y=( 4)^2-8( 4)+12

   ▼

   Simplify right-hand side

   CalcPow

   Calculate power

   y=16-8(4)+12

   Multiply

   Multiply

   y=16-32+12

   AddSubTerms

   Add and subtract terms

   y=- 4

   We found the y-coordinate, and now we know that the vertex is (4,- 4).

   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 (0,12). 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.

   

   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 6, approximately.