Math / Algebra

QuestionFind all $y$ -intercepts and $x$ -intercepts of the graph of the function. $f(x)=2 x^{2}-15 x+18$
If there is more than one answer, separate them with commas. Click on "None" if applicable.

Studdy Solution

STEP 1

1. The function is a quadratic function of the form $f(x) = ax^2 + bx + c$ .
2. The $y$ -intercept occurs where $x = 0$ .
3. The $x$ -intercepts occur where $f(x) = 0$ .

STEP 2

1. Find the $y$ -intercept.
2. Find the $x$ -intercepts by solving $f(x) = 0$ .

STEP 3

To find the $y$ -intercept, substitute $x = 0$ into the function $f(x)$ .
$f(0) = 2(0)^2 - 15(0) + 18 = 18$
Thus, the $y$ -intercept is at the point $(0, 18)$ .

STEP 4

To find the $x$ -intercepts, solve the equation $f(x) = 0$ .
$2x^2 - 15x + 18 = 0$

STEP 5

Use the quadratic formula to solve for $x$ , where $a = 2$ , $b = -15$ , and $c = 18$ .
The quadratic formula is:
$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$
Substitute the values of $a$ , $b$ , and $c$ .
$x = \frac{-(-15) \pm \sqrt{(-15)^2 - 4 \cdot 2 \cdot 18}}{2 \cdot 2}$
$x = \frac{15 \pm \sqrt{225 - 144}}{4}$
$x = \frac{15 \pm \sqrt{81}}{4}$
$x = \frac{15 \pm 9}{4}$

STEP 6

Calculate the two possible values for $x$ .
$x = \frac{15 + 9}{4} = \frac{24}{4} = 6$
$x = \frac{15 - 9}{4} = \frac{6}{4} = \frac{3}{2}$
Thus, the $x$ -intercepts are at the points $(6, 0)$ and $\left(\frac{3}{2}, 0\right)$ .
The $y$ -intercept is $(0, 18)$ and the $x$ -intercepts are $(6, 0)$ and $\left(\frac{3}{2}, 0\right)$ .

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QuestionFind all $y$ -intercepts and $x$ -intercepts of the graph of the function. $f(x)=2 x^{2}-15 x+18$
If there is more than one answer, separate them with commas. Click on "None" if applicable.

Studdy Solution

STEP 1

1. The function is a quadratic function of the form $f(x) = ax^2 + bx + c$ .
2. The $y$ -intercept occurs where $x = 0$ .
3. The $x$ -intercepts occur where $f(x) = 0$ .

STEP 2

1. Find the $y$ -intercept.
2. Find the $x$ -intercepts by solving $f(x) = 0$ .

STEP 3

To find the $y$ -intercept, substitute $x = 0$ into the function $f(x)$ .
$f(0) = 2(0)^2 - 15(0) + 18 = 18$
Thus, the $y$ -intercept is at the point $(0, 18)$ .

STEP 4

To find the $x$ -intercepts, solve the equation $f(x) = 0$ .
$2x^2 - 15x + 18 = 0$

STEP 5

STEP 6

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QuestionFind all yyy-intercepts and xxx-intercepts of the graph of the function. f(x)=2x2−15x+18f(x)=2 x^{2}-15 x+18f(x)=2x2−15x+18If there is more than one answer, separate them with commas. Click on "None" if applicable.

Studdy Solution

STEP 1

1. The function is a quadratic function of the form f(x)=ax2+bx+c f(x) = ax^2 + bx + c f(x)=ax2+bx+c.2. The y y y-intercept occurs where x=0 x = 0 x=0.3. The x x x-intercepts occur where f(x)=0 f(x) = 0 f(x)=0.

STEP 2

1. Find the y y y-intercept.2. Find the x x x-intercepts by solving f(x)=0 f(x) = 0 f(x)=0.

STEP 3

To find the y y y-intercept, substitute x=0 x = 0 x=0 into the function f(x) f(x) f(x).f(0)=2(0)2−15(0)+18=18 f(0) = 2(0)^2 - 15(0) + 18 = 18 f(0)=2(0)2−15(0)+18=18Thus, the y y y-intercept is at the point (0,18) (0, 18) (0,18).

STEP 4

To find the x x x-intercepts, solve the equation f(x)=0 f(x) = 0 f(x)=0.2x2−15x+18=0 2x^2 - 15x + 18 = 0 2x2−15x+18=0

STEP 5

STEP 6

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QuestionFind all $y$ -intercepts and $x$ -intercepts of the graph of the function. $f(x)=2 x^{2}-15 x+18$
If there is more than one answer, separate them with commas. Click on "None" if applicable.

1. The function is a quadratic function of the form $f(x) = ax^2 + bx + c$ .
2. The $y$ -intercept occurs where $x = 0$ .
3. The $x$ -intercepts occur where $f(x) = 0$ .

1. Find the $y$ -intercept.
2. Find the $x$ -intercepts by solving $f(x) = 0$ .

To find the $y$ -intercept, substitute $x = 0$ into the function $f(x)$ .
$f(0) = 2(0)^2 - 15(0) + 18 = 18$
Thus, the $y$ -intercept is at the point $(0, 18)$ .

To find the $x$ -intercepts, solve the equation $f(x) = 0$ .
$2x^2 - 15x + 18 = 0$