Problem $25$ . $10$ $\newline$ For the function $f(x)=x^{2}-6 x+9$ $\newline$ Determine the coordinates of the $x$ -intercepts. $\newline$ Determine the coordinates of the $y$ -intercept. $\newline$ Determine the equation of the axis of symmetry. $\newline$ Determine the coordinates of the vertex. $\newline$ Sketch a graph of the function, label the $x$ and $y$ axes with an appropriate scale

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Algebra 2

Characteristics of quadratic functions: equations

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Q. Problem

25

10

\newline

For the function

f(x)=x^{2}-6 x+9

\newline

Determine the coordinates of the

x

-intercepts.

\newline

Determine the coordinates of the

y

-intercept.

\newline

Determine the equation of the axis of symmetry.

\newline

Determine the coordinates of the vertex.

\newline

Sketch a graph of the function, label the

x

and

y

axes with an appropriate scale

Find x-intercepts: Step $1$ : Determine the x-intercepts of the function $f(x) = x^2 - 6x + 9$ . $\newline$ To find the x-intercepts, set $f(x) = 0$ and solve for $x$ . $\newline$ $0 = x^2 - 6x + 9$ $\newline$ Using the quadratic formula, $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ , where $a = 1$ , $b = -6$ , and $c = 9$ : $\newline$ $x = \frac{6 \pm \sqrt{(-6)^2 - 4\cdot1\cdot9}}{2\cdot1}$ $\newline$ $x = \frac{6 \pm \sqrt{36 - 36}}{2}$ $\newline$ $f(x) = 0$ $0$ $\newline$ $f(x) = 0$ $1$ $\newline$ $f(x) = 0$ $2$ $\newline$ Thus, the x-intercept is at $f(x) = 0$ $3$ .
Find y-intercept: Step $2$ : Determine the y-intercept of the function $f(x) = x^2 - 6x + 9$ . $\newline$ To find the y-intercept, set $x = 0$ and solve for $f(x)$ . $\newline$ $f(0) = 0^2 - 6\cdot0 + 9$ $\newline$ $f(0) = 9$ $\newline$ Thus, the y-intercept is at $(0,9)$ .
Axis of symmetry: Step $3$ : Determine the equation of the axis of symmetry. $\newline$ The axis of symmetry for a parabola in the form $y = ax^2 + bx + c$ is given by $x = -\frac{b}{2a}$ . $\newline$ For $f(x) = x^2 - 6x + 9$ , $a = 1$ and $b = -6$ : $\newline$ $x = -(-6)/(2\cdot1)$ $\newline$ $x = \frac{6}{2}$ $\newline$ $x = 3$ $\newline$ Thus, the axis of symmetry is $x = 3$ .
Find vertex: Step $4$ : Determine the coordinates of the vertex. $\newline$ The vertex of a parabola $y = ax^2 + bx + c$ is at the point $(x = -\frac{b}{2a}, f(x))$ . $\newline$ Using the axis of symmetry $x = 3$ : $\newline$ $f(3) = 3^2 - 6\cdot3 + 9$ $\newline$ $f(3) = 9 - 18 + 9$ $\newline$ $f(3) = 0$ $\newline$ Thus, the vertex is at $(3,0)$ .
Sketch graph: Step $5$ : Sketch a graph of the function. $\newline$ Plot the points found: x-intercept $(3,0)$ , y-intercept $(0,9)$ , and vertex $(3,0)$ . $\newline$ Draw a parabola opening upwards passing through these points. $\newline$ Label the $x$ and $y$ axes with an appropriate scale, typically $1$ unit per mark.