Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Find an expression for the function whose graph is the given curve. The line segment joining the points (-5, 10) and (7,-10).

Short Answer

Expert verified

The expression for the function is\({\bf{y = - }}\frac{{\bf{5}}}{{\bf{3}}}{\bf{x - }}\frac{{{\bf{16}}}}{{\bf{3}}}\).

Step by step solution

01

Determine the formula for the curve passing through different points

The equation of a line passing through the two points\(\left( {{x_1},{y_1}} \right)\)and\(\left( {{x_2},{y_2}} \right)\)is given as:

\(y - {y_1} = \frac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}\left( {x - {x_1}} \right)\)

02

Determine the expression of the curve passing through two points

The given two points are\(\left( { - 5,10} \right)\)and\(\left( {7, - 10} \right)\).

Substitute all the values in the above equation.

\(\begin{array}{c}y - 10 = \frac{{ - 10 - 10}}{{7 - \left( { - 5} \right)}}\left( {x - \left( { - 5} \right)} \right)\\y - 3 = \frac{{ - 20}}{{12}}\left( {x + 5} \right)\\y = - \frac{5}{3}x - \frac{{16}}{3}\end{array}\)

Therefore, the expression for the function is\(y = - \frac{5}{3}x - \frac{{16}}{3}\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free