QuestionFor the function $f(x)=x^{2}+6x$ , find if it opens up or down, and determine the vertex and intercepts.

Studdy Solution

STEP 1

Assumptions1. The quadratic function is $f(x)=x^{}+6 x$ . We need to determine the direction of the opening of the graph, the vertex, the axis of symmetry, the $y$ -intercept, and the $x$ -intercepts

STEP 2

The general form of a quadratic function is $f(x)=ax^{2}+bx+c$ . The sign of the coefficient $a$ determines whether the graph of the function opens up or down. If $a>0$ , the graph opens up. If $a<0$ , the graph opens down.

STEP 3

In our function $f(x)=x^{2}+6 x$ , the coefficient $a$ is1, which is greater than0. Therefore, the graph of the function opens up.

STEP 4

The vertex of a quadratic function given in the form $f(x)=ax^{2}+bx+c$ is at the point $(-\frac{b}{2a}, f(-\frac{b}{2a}))$ .

STEP 5

For our function $f(x)=x^{2}+ x$ , $a=1$ and $b=$ . So, the $x$ -coordinate of the vertex is $-\frac{b}{2a} = -\frac{}{2 \times1} = -3$ .

STEP 6

To find the $y$ -coordinate of the vertex, we substitute $x=-3$ into the function $f(x)$ .
$f(-3) = (-3)^{2}+6(-3)$

STEP 7

Calculate the value of $f(-3)$ .
$f(-3) =9 -18 = -9$

STEP 8

So, the vertex of the function is at the point $(-3, -)$ .

STEP 9

The axis of symmetry of a quadratic function given in the form $f(x)=ax^{2}+bx+c$ is the line $x=-\frac{b}{2a}$ .

STEP 10

For our function $f(x)=x^{2}+6 x$ , the axis of symmetry is the line $x=-3$ .

STEP 11

The $y$ -intercept of a function is the point where the graph of the function intersects the $y$ -axis. This occurs when $x=0$ .

STEP 12

Substitute $x=0$ into the function $f(x)$ to find the $y$ -intercept.
$f(0) =0^{2}+6(0)$

STEP 13

Calculate the value of $f(0)$ .
$f(0) =0$

STEP 14

So, the $y$ -intercept of the function is at the point $(0,0)$ .

STEP 15

The $x$ -intercepts of a function are the points where the graph of the function intersects the $x$ -axis. This occurs when $f(x)=0$ .

STEP 16

Set $f(x)$ equal to0 and solve for $x$ to find the $x$ -intercepts.
$0 = x^{2}+6x$

STEP 17

Factor the equation to solve for $x$ .
$0 = x(x+6)$

STEP 18

Set each factor equal to0 and solve for $x$ .
$x=0, x+6=0$

STEP 19

olve for $x$ .
$x=, x=-6$

STEP 20

So, the $x$ -intercepts of the function are at the points $(0,0)$ and $(-6,0)$ .

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