Let
f
(
x
)
=
x
2
\displaystyle {f{{\left({x}\right)}}}={x}^{{2}}
f
(
x
)
=
x
2
and
g
(
x
)
=
(
x
−
10
)
2
+
14.
\displaystyle {g{{\left({x}\right)}}}={\left({x}-{10}\right)}^{{2}}+{14}.
g
(
x
)
=
(
x
−
10
)
2
+
14
.
There is one line with positive slope that is tangent to both of the parabolas
y
=
f
(
x
)
\displaystyle {y}={f{{\left({x}\right)}}}
y
=
f
(
x
)
and
y
=
g
(
x
)
\displaystyle {y}={g{{\left({x}\right)}}}
y
=
g
(
x
)
simultaneously.
Find the equation of the line.
y =
Preview
Question 6
Submit
Try a similar question
License
[more..]
\displaystyle
Enter your answer as an expression. Example: 3x^2+1, x/5, (a+b)/c
Be sure your variables match those in the question
