Respuesta :
We are given the following:
- parabola passes to both (1,0) and (0,1)
- slope at x = 1 is 4 from the equation of the tangent line
First, we figure out the value of c or the y intercept, we use the second point (0, 1) and substitute to the equation of the parabola. When x = 0, y = 1. So, c should be equal to 1. The parabola is y = ax^2 + bx + 1
Now, we can substitute the point (1,0) into the equation,
0 = a(1)^2 + b(1) + 1
0 = a + b + 1
a + b = -1
The slope at x = 1 is equal to 4 which is equal to the first derivative of the equation.
We take the derivative of the equation ,
y = ax^2 + bx + 1
y' = 2ax + b
x = 1, y' = 2
4 = 2a(1) + b
4 = 2a + b
So, we have two equations and two unknowns,
2a + b = 4
a + b = -1
Solving simultaneously,
a = 5
b = -6
Therefore, the eqution of the parabola is y = 5x^2 - 6x + 1 .
- parabola passes to both (1,0) and (0,1)
- slope at x = 1 is 4 from the equation of the tangent line
First, we figure out the value of c or the y intercept, we use the second point (0, 1) and substitute to the equation of the parabola. When x = 0, y = 1. So, c should be equal to 1. The parabola is y = ax^2 + bx + 1
Now, we can substitute the point (1,0) into the equation,
0 = a(1)^2 + b(1) + 1
0 = a + b + 1
a + b = -1
The slope at x = 1 is equal to 4 which is equal to the first derivative of the equation.
We take the derivative of the equation ,
y = ax^2 + bx + 1
y' = 2ax + b
x = 1, y' = 2
4 = 2a(1) + b
4 = 2a + b
So, we have two equations and two unknowns,
2a + b = 4
a + b = -1
Solving simultaneously,
a = 5
b = -6
Therefore, the eqution of the parabola is y = 5x^2 - 6x + 1 .
Applying the conditions, the equation of the parabola is given by:
[tex]y = 5x^2 - 6x + 1[/tex]
The standard equation of a parabola is given by:
[tex]y = ax^2 + bx + c[/tex]
In this question, it passes through (0,1), thus when [tex]x = 0, y = 1[/tex], which means that:
[tex]1 = a(0)^2 + b(0) + c[/tex]
[tex]c = 1[/tex]
Then:
[tex]y = ax^2 + bx + 1[/tex].
Tangent to the line y = 4x - 4 at (1,0), which means that, since the slope of the line of 4 is the derivative at x = 1.
[tex]2a + b = 4[/tex]
The parabola also passes through (1,0), thus:
[tex]a + b + 1 = 0[/tex]
[tex]a + b = -1[/tex]
Solving the system:
[tex]2a + b = 4[/tex]
[tex]a + b = -1[/tex]
Subtracting the first equation by the second:
[tex]2a - a + b - b = 4 - (-1)[/tex]
[tex]a = 5[/tex]
Then:
[tex]b = -1 - a = -1 - 5 = -6[/tex]
Thus, the equation is:
[tex]y = 5x^2 - 6x + 1[/tex]
A similar problem is given at https://brainly.com/question/22426360
