Kaeden Hoffman

Answered

2022-07-06

With this information I am to find the linear equation: f(−5)=−4 and f(5)=2

The solution is provided as $y=\frac{3}{5}x-1$ however I arrived at $y=\frac{7}{10}x-\frac{3}{2}$

Here is my working:

Using these points, find the equation: (-5,-4)(5, 2)

the slope m: $m=\frac{y1-y}{x1-x}$=$\frac{2+5}{5+5}$=$\frac{7}{10}$

Now that I have m, plug the values of one of the pairs into the linear function form to find b: $y=mx+b$

$2=\frac{7}{10}(5)+b$

$\frac{7}{10}(5)+b=2$

$\frac{7}{2}+b=2$ #7/10 * 5 = 7/2

$b=2-\frac{7}{2}$

$b=-\frac{3}{2}$

$y=\frac{7}{10}x+\frac{3}{2}$

Where did I go wrong and how can I arrive at $y=\frac{3}{5}x-1$?

The solution is provided as $y=\frac{3}{5}x-1$ however I arrived at $y=\frac{7}{10}x-\frac{3}{2}$

Here is my working:

Using these points, find the equation: (-5,-4)(5, 2)

the slope m: $m=\frac{y1-y}{x1-x}$=$\frac{2+5}{5+5}$=$\frac{7}{10}$

Now that I have m, plug the values of one of the pairs into the linear function form to find b: $y=mx+b$

$2=\frac{7}{10}(5)+b$

$\frac{7}{10}(5)+b=2$

$\frac{7}{2}+b=2$ #7/10 * 5 = 7/2

$b=2-\frac{7}{2}$

$b=-\frac{3}{2}$

$y=\frac{7}{10}x+\frac{3}{2}$

Where did I go wrong and how can I arrive at $y=\frac{3}{5}x-1$?

Answer & Explanation

amanhantmk

Expert

2022-07-07Added 17 answers

Probably you made a mistake in the step below:

“Using these points, find the equation: (-5,-4)(5, 2)”

You should calculate 2-(-4)

“Using these points, find the equation: (-5,-4)(5, 2)”

You should calculate 2-(-4)

therightwomanwf

Expert

2022-07-08Added 4 answers

Point-slope form for the line might be a bit easier:

$y-{y}_{1}=m(x-{x}_{1})$

$m={\displaystyle \frac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}}}={\displaystyle \frac{2-(-4)}{5-(-5)}}={\displaystyle \frac{3}{5}}$

$y-2={\displaystyle \frac{3}{5}}(x-5)$

$y={\displaystyle \frac{3}{5}}x-1$

$y-{y}_{1}=m(x-{x}_{1})$

$m={\displaystyle \frac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}}}={\displaystyle \frac{2-(-4)}{5-(-5)}}={\displaystyle \frac{3}{5}}$

$y-2={\displaystyle \frac{3}{5}}(x-5)$

$y={\displaystyle \frac{3}{5}}x-1$

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