Find the slope of \(\displaystyle y=3x+10\).  The slope is 3.

Substitute \(\displaystyle x=2\) to determine the y-value.

\(\displaystyle y=3x+10=3(2)+10=16\)

The point is \(\displaystyle (2,16)\).

Use the slope-intercept formula to find the y-intercept, given the point and slope.

\(\displaystyle y=mx+b\)

Substitute the point and the slope.

\(\displaystyle 16=3(2)+b\)

\(\displaystyle b=10\)

Substitute the y-intercept and slope back to the slope-intercept formula.

The correct answer is:  \(\displaystyle y=3x+10\)

Write \(\displaystyle 2y=3x+4\) in slope-intercept form \(\displaystyle y=mx+b\) and determine the slope.

Rearranging our equation to be in slope-intercept form we get:

\(\displaystyle y=\frac{3}{2}x+2\).

The slope is our \(\displaystyle m\) value which is \(\displaystyle \frac{3}{2}\).

Substitute the slope and the point to the slope-intercept form.

\(\displaystyle y=mx+b\)

\(\displaystyle 4=\frac{3}{2}(2)+b\)

\(\displaystyle b=1\)

Substitute the slope and y-intercept to find our final equation.

\(\displaystyle y=\frac{3}{2}x+1\)

Rewrite \(\displaystyle 5y-3x+2=0\) in slope-intercept form to determine the slope. Remember slope-intercept form is \(\displaystyle y=mx+b\).

Therefore, the equation becomes,

\(\displaystyle 5y=3x-2\)

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

The slope is the \(\displaystyle m\) value of the function thus, it is \(\displaystyle \frac{3}{5}\).

Substitute \(\displaystyle x=0\) back to the original equation to find the value of \(\displaystyle y\).

\(\displaystyle 5y-3(0)+2=0\)

\(\displaystyle 5y+2=0\)

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

Substitute the point \(\displaystyle \left(0,-\frac{2}{5}\right)\) and the slope of the line into the slope-intercept equation.

\(\displaystyle y=mx+b\)

\(\displaystyle -\frac{2}{5}=\left(\frac{3}{5}\right)(0)+b\)

\(\displaystyle b=-\frac{2}{5}\)

Substitute the point and the slope back in to the slope-intercept formula.

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

The circle's center is \(\displaystyle (0,3)\). The tangent line will be perpendicular to the line going through the points \(\displaystyle (0,3)\) and \(\displaystyle (4, -2)\), so it will be helpful to know the slope of this line:

\(\displaystyle \frac{ \Delta y }{ \Delta x } = \frac{ 3 - - 2 }{ 0 - 4 } = \frac{ 5 }{ -4 }\)

Since the tangent line is perpendicular, its slope is \(\displaystyle \frac{4}{5 }\)

To write the equation in the form \(\displaystyle y = mx + b\), we need to solve for "b," the y-intercept. We can plug in the slope for "m" and the coordinates of the point \(\displaystyle (4, -2)\) for x and y:

\(\displaystyle -2 = \frac{4}{5}(4) + b\)

\(\displaystyle -2 = \frac{16} { 5 } + b\)

\(\displaystyle -\frac{10}{5} = \frac{16} {5} + b\)

\(\displaystyle -\frac{26}{5} = b\)

The equation is \(\displaystyle y = \frac{4}{5} x - \frac{26} {5 }\)

The tangent to a circle at a given point is perpendicular to the radius that has the center and the given point as its endpoints.

The circle has its center at origin \(\displaystyle (0,0)\); since this and \(\displaystyle (4,9)\) are the endpoints of the radius - and the line that includes this radius includes both points - its slope can be found by setting \(\displaystyle x_{1} = y_{1} = 0 , x_{2}= 4, y_{2} = 9\) in the following slope formula:

\(\displaystyle m = \frac{y_{2}-y_{1}}{x_{2}-x_{1}}\)

\(\displaystyle m = \frac{9-0}{4-0} = \frac{9 }{4 }\)

The tangent line, being perpendicular to this radius, has as its slope the opposite of the reciprocal of this, which is \(\displaystyle m = -\frac{4}{9}\). Since the tangent line includes point \(\displaystyle (4, 9)\), set \(\displaystyle m = -\frac{4}{9}, x_{1} = 4, y_{1} = 9\) in the point-slope formula and simplify:

\(\displaystyle y - y_{1} = m (x-x_{1})\)

\(\displaystyle y - 9= -\frac{4}{9} (x-4)\)

\(\displaystyle y - 9= -\frac{4}{9} \cdot x- \left (-\frac{4}{9} \right ) \cdot 4\)

\(\displaystyle y - 9= -\frac{4}{9} x+ \frac{16}{9}\)

\(\displaystyle y - 9+ 9 = -\frac{4}{9} x+ \frac{16}{9} + 9\)

\(\displaystyle y = -\frac{4}{9} x+ \frac{16}{9} + \frac{81}{9}\)

\(\displaystyle y = -\frac{4}{9} x+ \frac{97}{9}\)

The center of this circle is \(\displaystyle (4,2)\)

Therefore, the radius with endpoint \(\displaystyle (10,-6)\) has slope  

\(\displaystyle \frac{2-(-6)}{4-10}= \frac{8}{-6}= -\frac{4}{3}\)

The tangent line at \(\displaystyle (10,-6)\) is perpendicular to this radius; therefore, its slope is the opposite of the reciprocal of \(\displaystyle -\frac{4}{3}\), or \(\displaystyle \frac{3}{4}\). 

Now use the point-slope formula with this slope and the point of tangency:

\(\displaystyle y-(-6)= \frac{3}{4} \left ( x-10\right )\)

\(\displaystyle y+6= \frac{3}{4}x-\frac{15}{2}\)

\(\displaystyle y= \frac{3}{4}x-\frac{27}{2}\)

To find the slope of the tangent line, first we must take the derivative of \(\displaystyle x^2-6\), giving us \(\displaystyle 2x\). Next we simply plug in our given x-value, which in this case is \(\displaystyle x=3\). This leaves us with a slope of \(\displaystyle 6\).