```Finding Intersections
• Equation of arc 1:
y  .0070 x  1.2349 x
2
• Equation of arc 2:
y  .0254 x  4.4856 x  136.9897
2
• To find intersections, use substitution:
» EQ ARC 1 = EQ ARC 2
Finding Intersections
 .0070 x 2  1.2349 x  .0254 x 2  4.4856 x  136.9897
 .0070 x  1.2349 x   .0070 x  1.2349 x
2
2
0  .0184 x  3.2507 x  136.9897
2
• I solved the system of equations so that one
side equaled 0. Now, I will use the
quadratic formula to find my x-values.
Finding Intersections
2
0  .0184 x  3.2507 x  136.9897
b
b  4ac
x

2a
2a
2
2
3
.
2507
 4(.0184)( 136.9897)
 3.2507
x

2(.0184)
2(.0184)
 3.2507
10.5671  10.0824
x

 .0368
.0368
Finding Intersections
 3.2507
.4846
x

 .0368
.0368
 3.2507 .6961
x

 .0368 .0368
x  88.3342 18.9168
x  107.2510
x  69.4174
Finding Intersections
• Now that we have found the x-values of the
intersections, substitute into an equation for
an arc to find the y-value.
• I find the non-translated one to be just a bit
easier.
Finding Intersections
x  107.2510
y  .0070 x  1.2349 x
x  69.4174
2
y  .0070(107.2510)  1.2349(107.2510)
2
y  .0070(11502.7770)  1.2349(107.2510)
y  80.5194  132.4443
y  51.9249
Finding Intersections
x

107
.
2510
2
y  .0070 x  1.2349 x
x  69.4174
y  .0070(69.4174)  1.2349(69.4174)
y  .0070(4818.7754)  1.2349(69.4174)
y  33.7314  85.7235
y  51.9921
2
I know that the y-values should be exactly the same since I have
the same line of symmetry. These are basically the same, where
the rounding I have done caused the difference. I’ll label the
y-value as the midpoint between the two solutions I got – 59.9585
Finding Intersections
```