What is the maximum value of c such that the graph of the parabola y = (1/3)x^2 has at most one point of intersection with the line y = x+c?
It is actually -3/4
To find the maximum value of c such that the graph of the parabola y = (1/3)x^2 has at most one point of intersection with the line y = x + c, we need to determine the conditions for intersection and analyze the possibilities.
Step 1: Set the equations equal to each other and solve for x:
(1/3)x^2 = x + c
Multiply both sides by 3 to eliminate the fraction:
x^2 = 3x + 3c
Step 2: Rewrite the equation in standard quadratic form:
x^2 - 3x - 3c = 0
Step 3: Use the discriminant to determine the number of intersection points:
The discriminant (D) of a quadratic equation ax^2 + bx + c = 0 is given by D = b^2 - 4ac.
If D > 0, the equation has two real roots and the parabola intersects the line at two points.
If D = 0, the equation has one real root and the parabola is tangent to the line at one point.
If D < 0, the equation has no real roots and the parabola does not intersect the line.
In this case, our quadratic equation is x^2 - 3x - 3c = 0, where a = 1, b = -3, and c = -3c. Therefore, the discriminant is:
D = (-3)^2 - 4(1)(-3c)
= 9 + 12c
Step 4: Apply the condition for at most one intersection point:
Based on the question, we want at most one intersection point. This means that the discriminant D should be less than or equal to zero:
D ≤ 0
Substituting the value of D, we get:
9 + 12c ≤ 0
Step 5: Solve for c:
Subtract 9 from both sides of the inequality:
12c ≤ -9
Divide both sides by 12:
c ≤ -9/12
Simplifying further, we have:
c ≤ -3/4
Therefore, the maximum value of c such that the graph of the parabola y = (1/3)x^2 has at most one point of intersection with the line y = x + c is c ≤ -3/4.
x+c=1/3 x^2
x^2-3x+3c=0
making it one solution,
x^2-3x+(3/2)^2=0
or 3c=(9/4) or 3/4
x-3x-9/4=0
(x-3/2)^2=0 single solution occurs at that x=3/2 and max c is 3/4
check my work and thinking.