Chapter 8.3, Problem 33E

To determine

To calculate: The vertex of the graph of the quadratic function $y=0.5x^2+\sqrt{2}x-3$ .

Answer to Problem 33E

The vertex of the graph of the quadratic function $y=0.5x^2+\sqrt{2}x-3$ is $\left(-1.414,-4\right)$ .

Explanation of Solution

Given information:

The quadratic function $y=0.5x^2+\sqrt{2}x-3$ .

Formula used:

The quadratic function of the form $y=a{x}^2+bx+c$ , the y -coordinate of the vertex is the maximum value of the function when $a<0$ and minimum value of the function when $a>0$ .

Calculation:

Consider the quadratic function $y=0.5x^2+\sqrt{2}x-3$ .

The steps to evaluate the vertex of the function is provided below,

Step 1: Press $\boxed{\text{Y}=}$ key.

Step 2: Enter the function $0.5x^2+\sqrt{2}x-3$ .

Step 3: Press $\boxed{\text{GRAPH}}$ key.

The result obtained is provided below,

  

Now, the graph open upward so the function has minimum value. Also recall that quadratic function of the form $y=a{x}^2+bx+c$ , the y -coordinate of the vertex is the maximum value of the function when $a<0$ and minimum value of the function when $a>0$ .

Here, $a=0.5$ .

Step 4: Press the $\boxed{\text{2nd}}$ key and then $\boxed{\text{TRACE}}$ key.

Step 5: Choose the third option that is minimum from the list.

Step 6: Enter the left bound as $-2.97$ , right bound as $0$ and guess as $-1.2$ where the vertex might lie with help of arrow key and press $\boxed{\text{ENTER}}$ key.

The result obtained is provided below,

  

The vertex of the graph of the quadratic function $y=0.5x^2+\sqrt{2}x-3$ is $\left(-1.414,-4\right)$ .