Find the x and y Intercepts f(x)=10x^2-6x-3

$f (x) = 10 x^{2} - 6 x - 3$

Step 1

Find the x-intercepts.

To find the x-intercept(s), substitute in $0$ for $y$ and solve for $x$ .

$0 = 10 x^{2} - 6 x - 3$

Solve the equation.

Rewrite the equation as $10 x^{2} - 6 x - 3 = 0$ .

$10 x^{2} - 6 x - 3 = 0$

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Substitute the values $a = 10$ , $b = - 6$ , and $c = - 3$ into the quadratic formula and solve for $x$ .

$\frac{6 \pm \sqrt{{(- 6)}^{2} - 4 \cdot (10 \cdot - 3)}}{2 \cdot 10}$

Simplify.

Simplify the numerator.

Raise $- 6$ to the power of $2$ .

$x = \frac{6 \pm \sqrt{36 - 4 \cdot 10 \cdot - 3}}{2 \cdot 10}$

Multiply $- 4 \cdot 10 \cdot - 3$ .

Multiply $- 4$ by $10$ .

$x = \frac{6 \pm \sqrt{36 - 40 \cdot - 3}}{2 \cdot 10}$

Multiply $- 40$ by $- 3$ .

$x = \frac{6 \pm \sqrt{36 + 120}}{2 \cdot 10}$

Add $36$ and $120$ .

$x = \frac{6 \pm \sqrt{156}}{2 \cdot 10}$

Rewrite $156$ as $2^{2} \cdot 39$ .

Factor $4$ out of $156$ .

$x = \frac{6 \pm \sqrt{4 (39)}}{2 \cdot 10}$

Rewrite $4$ as $2^{2}$ .

$x = \frac{6 \pm \sqrt{2^{2} \cdot 39}}{2 \cdot 10}$

Pull terms out from under the radical.

$x = \frac{6 \pm 2 \sqrt{39}}{2 \cdot 10}$

Multiply $2$ by $10$ .

$x = \frac{6 \pm 2 \sqrt{39}}{20}$

Simplify $\frac{6 \pm 2 \sqrt{39}}{20}$ .

$x = \frac{3 \pm \sqrt{39}}{10}$

Simplify the expression to solve for the $+$ portion of the $\pm$ .

Simplify the numerator.

Raise $- 6$ to the power of $2$ .

$x = \frac{6 \pm \sqrt{36 - 4 \cdot 10 \cdot - 3}}{2 \cdot 10}$

Multiply $- 4 \cdot 10 \cdot - 3$ .

Multiply $- 4$ by $10$ .

$x = \frac{6 \pm \sqrt{36 - 40 \cdot - 3}}{2 \cdot 10}$

Multiply $- 40$ by $- 3$ .

$x = \frac{6 \pm \sqrt{36 + 120}}{2 \cdot 10}$

Add $36$ and $120$ .

$x = \frac{6 \pm \sqrt{156}}{2 \cdot 10}$

Rewrite $156$ as $2^{2} \cdot 39$ .

Factor $4$ out of $156$ .

$x = \frac{6 \pm \sqrt{4 (39)}}{2 \cdot 10}$

Rewrite $4$ as $2^{2}$ .

$x = \frac{6 \pm \sqrt{2^{2} \cdot 39}}{2 \cdot 10}$

Pull terms out from under the radical.

$x = \frac{6 \pm 2 \sqrt{39}}{2 \cdot 10}$

Multiply $2$ by $10$ .

$x = \frac{6 \pm 2 \sqrt{39}}{20}$

Simplify $\frac{6 \pm 2 \sqrt{39}}{20}$ .

$x = \frac{3 \pm \sqrt{39}}{10}$

Change the $\pm$ to $+$ .

$x = \frac{3 + \sqrt{39}}{10}$

Simplify the expression to solve for the $-$ portion of the $\pm$ .

Simplify the numerator.

Raise $- 6$ to the power of $2$ .

$x = \frac{6 \pm \sqrt{36 - 4 \cdot 10 \cdot - 3}}{2 \cdot 10}$

Multiply $- 4 \cdot 10 \cdot - 3$ .

Multiply $- 4$ by $10$ .

$x = \frac{6 \pm \sqrt{36 - 40 \cdot - 3}}{2 \cdot 10}$

Multiply $- 40$ by $- 3$ .

$x = \frac{6 \pm \sqrt{36 + 120}}{2 \cdot 10}$

Add $36$ and $120$ .

$x = \frac{6 \pm \sqrt{156}}{2 \cdot 10}$

Rewrite $156$ as $2^{2} \cdot 39$ .

Factor $4$ out of $156$ .

$x = \frac{6 \pm \sqrt{4 (39)}}{2 \cdot 10}$

Rewrite $4$ as $2^{2}$ .

$x = \frac{6 \pm \sqrt{2^{2} \cdot 39}}{2 \cdot 10}$

Pull terms out from under the radical.

$x = \frac{6 \pm 2 \sqrt{39}}{2 \cdot 10}$

Multiply $2$ by $10$ .

$x = \frac{6 \pm 2 \sqrt{39}}{20}$

Simplify $\frac{6 \pm 2 \sqrt{39}}{20}$ .

$x = \frac{3 \pm \sqrt{39}}{10}$

Change the $\pm$ to $-$ .

$x = \frac{3 - \sqrt{39}}{10}$

The final answer is the combination of both solutions.

$x = \frac{3 + \sqrt{39}}{10}, \frac{3 - \sqrt{39}}{10}$

x-intercept(s) in point form.

x-intercept(s): $(\frac{3 + \sqrt{39}}{10}, 0), (\frac{3 - \sqrt{39}}{10}, 0)$

Step 2

Find the y-intercepts.

To find the y-intercept(s), substitute in $0$ for $x$ and solve for $y$ .

$y = 10 {(0)}^{2} - 6 \cdot 0 - 3$

Solve the equation.

Remove parentheses.

$y = 10 \cdot 0^{2} - 6 \cdot 0 - 3$

Remove parentheses.

$y = 10 {(0)}^{2} - 6 \cdot 0 - 3$

Simplify $10 {(0)}^{2} - 6 \cdot 0 - 3$ .

Simplify each term.

Raising $0$ to any positive power yields $0$ .

$y = 10 \cdot 0 - 6 \cdot 0 - 3$

Multiply $10$ by $0$ .

$y = 0 - 6 \cdot 0 - 3$

Multiply $- 6$ by $0$ .

$y = 0 + 0 - 3$

Simplify by adding and subtracting.

Add $0$ and $0$ .

$y = 0 - 3$

Subtract $3$ from $0$ .

$y = - 3$

y-intercept(s) in point form.

y-intercept(s): $(0, - 3)$

Step 3

List the intersections.

x-intercept(s): $(\frac{3 + \sqrt{39}}{10}, 0), (\frac{3 - \sqrt{39}}{10}, 0)$

y-intercept(s): $(0, - 3)$

Step 4

Find the x and y Intercepts f(x)=10x^2-6x-3

Related Questions

Support LCMGCF.com by disabling your adblocker.