Answer:
c=6.25
Step-by-step explanation:
If a line that is flat (has a slope of zero) hits only one point of the parabola, it means that point is the vertex of the parabola.
In this case your equation is ---
y= –x^2 + 5x
--- and you just need to write in vertex for the equation:
y=a(x–h)^2 + k
y= –(x–2.5)^2 + 6.25
The vertex is (h,k) or (2.5,6.25)
Since the equation is "y=", k=c
or c=6.25
Thank You.
In this question:
Doing this, we find that the value is: [tex]c = \frac{25}{4}[/tex]
Solving a quadratic equation:
Given a second order polynomial expressed by the following equation:
[tex]ax^{2} + bx + c, a\neq0[/tex].
This polynomial has roots [tex]x_{1}, x_{2}[/tex] such that [tex]ax^{2} + bx + c = a(x - x_{1})*(x - x_{2})[/tex], given by the following formulas:
[tex]x_{1} = \frac{-b + \sqrt{\Delta}}{2*a}[/tex]
[tex]x_{2} = \frac{-b - \sqrt{\Delta}}{2*a}[/tex]
[tex]\Delta = b^{2} - 4ac[/tex]
Line: y = c
Parabola: y = -x^ 2 +5x.
The intersection is at the point:
[tex]-x^2 + 5x = c[/tex]
[tex]-x^2 + 5x - c = 0[/tex]
[tex]x^2 - 5x + c = 0[/tex]
Which is a quadratic equation with [tex]a = 1, b = -5[/tex].
The line and the parabola intersect at exactly one point, which means that the quadratic equation has one solution, that is, [tex]\Delta = 0[/tex].
Since:
[tex]\Delta = b^{2} - 4ac[/tex]
It means that:
[tex]b^2 - 4ac = 0[/tex]
We use this to find the value of c.
[tex](-5)^2 - 4(1)(c) = 0[/tex]
[tex]4c = 25[/tex]
[tex]c = \frac{25}{4}[/tex]
The value is: [tex]c = \frac{25}{4}[/tex]
A similar question is found at https://brainly.com/question/15211279
