Exercise 13 Page 534

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-5 ⇔ y=1x^2+2x+(- 5) We can see that a=1, b=2, and c=- 5. 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

   QuotOne

   a/a=1

   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-5

   Substitute

   x= - 1

   y=( - 1)^2+2( - 1)-5

   ▼

   Simplify right-hand side

   NegBaseToPosBase

   (- a)^2=a^2

   y=1+2(- 1)-5

   MultPosNeg

   a(- b)=- a * b

   y=1-2-5

   SubTerms

   Subtract terms

   y=- 6

   We found the y-coordinate, and now we know that the vertex is (- 1,- 6).

   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,- 5). 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 zeros of the function. We can see that the parabola intercepts the x-axis at x=- 3.4 and x=1.4.