A quadratic function is defined by
f(x)= 3x^2+4x-2. A linear function is defined by g(x)= mx-5. what values(s) of the slope of the line would make it a tangent to the parabola.
How can i get 6*x+4?
Slope of any function is first derivate of that function.
Quadratic function in your case have first dervate:
3*2*x+4=6*x+4
Linear function in your case have first dervate:
m
Line will be tangent of quadratic parabola when first derivates is egual.
m=6*x+4
m*x-5=(6*x+4)*x-5=6*x^2+4*x-5
3*x^2+4*x-2=6*x^2+4*x-5
3*x^2+4*x-6*x^2-4*x=-5+2
-3*x^2=-3 Multiply with (-1)
3*x^2=3 Divided with 3
x^2=1
x=+OR-(1)
x1=-1 , x2=1
Slope for x1=6*x1+4=6*(-1)+4=-6+4=-2
Slope for x2=6*1+4=6+4=10
S1=-2 S2=10
To find the value(s) of the slope (m) that would make the line a tangent to the parabola, we need to find the point(s) of tangency first.
Since a tangent line to a curve will have the same slope as the curve at the point of tangency, we can equate the slopes of the quadratic function and the linear function to find the values of m.
First, let's find the derivative of the quadratic function f(x) to determine its slope at any given point:
f(x) = 3x^2 + 4x - 2
Differentiating both sides with respect to x:
f'(x) = d/dx (3x^2 + 4x - 2)
= 6x + 4
Now, we set the derivative equal to the slope of the linear function g(x), which is m:
6x + 4 = m
To find the x-coordinate(s) of the point(s) of tangency, we set the quadratic function equal to the linear function:
3x^2 + 4x - 2 = mx - 5
Rearranging the equation to set it equal to zero:
3x^2 + (4 - m)x + (3 - 5) = 0
3x^2 + (4 - m)x - 2 = 0
Now, for a line to be tangent to a parabola, there should be only one solution for x. In other words, the quadratic equation should have a discriminant (b^2 - 4ac) equal to zero.
To find the discriminant, we substitute the coefficients into the quadratic formula:
Discriminant = (4 - m)^2 - 4(3)(-2)
= (16 - 8m + m^2) + 24
= m^2 - 8m + 40
Since the discriminant should be equal to zero, we set it equal to zero and solve for m:
m^2 - 8m + 40 = 0
Using the quadratic formula:
m = (-b ± √(b^2 - 4ac)) / (2a)
We have a = 1, b = -8, and c = 40:
m = (-(-8) ± √((-8)^2 - 4(1)(40))) / (2(1))
m = (8 ± √(64 - 160)) / 2
m = (8 ± √(-96)) / 2
Since the discriminant (-96) is negative, the quadratic equation does not have real solutions. Thus, there are no values of m that would make the line a tangent to the parabola given by f(x) = 3x^2 + 4x - 2.
To find the slope of the line that would make it a tangent to the parabola, we need to determine the point of tangency between the line and the parabola. The slope of the line at that point will be equal to the slope of the tangent line.
Step 1: Find the derivative of the quadratic function, f(x), to determine the slope of the tangent line at any given point.
The derivative of f(x) = 3x^2 + 4x - 2 can be found by applying the power rule of differentiation:
f'(x) = 6x + 4
Step 2: Set the derivative equal to the slope of the linear function, g(x), since they are both tangent lines.
6x + 4 = m
Step 3: Solve for the value(s) of x.
To find the x-coordinate(s) of the point(s) of tangency, we set the quadratic equation equal to the equation of the linear function and solve for x:
3x^2 + 4x - 2 = mx - 5
Rewriting the equation in standard form:
3x^2 + (4 - m)x + (5 - 2) = 0
Using the quadratic formula:
x = [-b ± √(b^2 - 4ac)] / (2a)
where a = 3, b = (4 - m), and c = (5 - 2).
To determine the values of x that will result in tangency, we need the discriminant (b^2 - 4ac) to be equal to zero. This means that the quadratic equation has exactly one real root, which corresponds to the point(s) of tangency.
Step 4: Set the discriminant equal to zero.
(4 - m)^2 - 4(3)(5 - 2) = 0
Simplifying the equation:
(4 - m)^2 - 12(5 - 2) = 0
Expanding and simplifying:
16 - 8m + m^2 - 60 + 24 = 0
Rearranging and combining like terms:
m^2 - 8m - 20 = 0
Step 5: Solve the resulting quadratic equation for the possible values of m.
Using the quadratic formula:
m = [-(-8) ± √((-8)^2 - 4(1)(-20))] / (2(1))
Simplifying:
m = (8 ± √(64 + 80)) / 2
m = (8 ± √144) / 2
m = (8 ± 12) / 2
This gives us two possibilities:
1. m = (8 + 12) / 2 = 10
2. m = (8 - 12) / 2 = -2
Therefore, the values of the slope of the line that would make it a tangent to the parabola are m = 10 and m = -2.