Chapter 9.1, Problem 59PPS

To determine

To Graph and Find:

The given function.

Find vertex using the TRACE feature.

Answer to Problem 59PPS

Vertex : (-1.25,-0.25)

Explanation of Solution

Given Information:

Given equation is :

  $y=4x^2+10x+6$

Concept Used :

• Quadratic Functions:
• Quadratic functions are non-linear and are of the form:

  $f(x)=a{x}^2+bx+c;a\ne 0$

This is the standard form of Quadratic function.

• If a > 0 ,
• The graph of $a{x}^2+bx+c$ opens upward.
• The lowest point on the graph is the minimum.
• If a < 0 ,
• The graph of $a{x}^2+bx+c$ opens downward.
• The highest point on the graph is the maximum.
• The maximum or the minimum is the vertex.

Calculation:

• Using the TI-83 Plus Calculator :

• Step 1 : Let Y1 = $4x^2+10x+6$

  

• Step 2 : Use the WINDOW option to adjust the viewing window.

  

• Step 3: Use the GRAPH option to get the graph of the function.

  

• Step 4: Use the TRACE option to move the cursor to the minimum point .

  

So, the vertex is at (-1.25,-0.25).