Answer:
y = 2x - 1
Step-by-step explanation:
A line y = mx + d is tangent to the parabola y = ax² + bx + c if the equation mx + d = ax² + bx + c has exactly one solution, and the discriminant of the quadratic equation is zero.
For function f(x) = x², the equation is:
[tex]mx + d = x^2[/tex]
Rearrange to the form ax² + bx + c = 0:
[tex]x^2 - mx - d = 0[/tex]
Therefore, a = 1, b = -m, and c = -d.
Substitute these values into the discriminant (b² - 4ac), set it to zero, and simplify:
[tex]\begin{aligned}(-m)^2 - 4(1)(-d) &= 0\\\\m^2+4d&=0\end{aligned}[/tex]
For function g(x) = (x - 3)² + 6, the equation is:
[tex]mx + d = (x - 3)^2 + 6[/tex]
Expand the right side, and rearrange to the form ax² + bx + c = 0:
[tex]\begin{aligned}mx + d &= (x - 3)^2 + 6\\\\mx+d&=x^2-6x+15\\\\x^2-6x-mx+15-d&=0\\\\x^2+(-6-m)x+(15-d)&=0\end{aligned}[/tex]
Therefore, a = 1, b = (-6 - m), and c = (15 - d).
Substitute these values into the discriminant (b² - 4ac), set it to zero, and simplify:
[tex]\begin{aligned}(-6-m)^2-4(1)(15-d)&=0\\\\36+12m+m^2-60+4d&=0\\\\12m-24+m^2+4d&=0\end{aligned}[/tex]
Now, we have created a system of equations in terms of the slope (m) and the y-intercept (d) of the line tangent to both parabolas.
[tex]\begin{cases}m^2 + 4d = 0\\m^2 + 12m - 24 + 4d = 0\end{cases}[/tex]
To find the slope (m), substitute the first equation into the second equation and solve for m:
[tex]\begin{aligned}12m-24+0&=0\\\\12m-24&=0\\\\12m&=24\\\\m&=2\end{aligned}[/tex]
Therefore, the slope of the tangent line is 2.
To find the y-intercept (d), substitute the value of m into the first equation and solve for d:
[tex]\begin{aligned}(2)^2+4d&=0\\\\4+4d&=0\\\\4d&=-4\\\\d&=-1\end{aligned}[/tex]
Therefore, the y-intercept of the tangent line is -1.
So, the equation of the line that is tangent to both of the parabolas of y = f(x) and y = g(x) simultaneously is:
[tex]\Large\boxed{\boxed{y=2x-1}}[/tex]
