A quadratic polynomial y = f(x) with absolute | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) Given $f(x) = ax^2 + bx + c$. Since the absolute term is $3$,$c = 3$. Since the leading coefficient is unity,$a = 1$. Thus,$f(x) = x^2 + bx + 3$.S

Vedclass · Accepted Answer

$(1, 9/4)$

Vedclass · Answer

(D) Given $f(x) = ax^2 + bx + c$. Since the absolute term is $3$,$c = 3$. Since the leading coefficient is unity,$a = 1$. Thus,$f(x) = x^2 + bx + 3$.Since the parabola is symmetric about $x = 1$,the vertex is at $x = 1$. The derivative $f'(x) = 2x + b$. Setting $f'(1) = 0$,we get $2(1) + b = 0$,so $b = -2$.Therefore,the polynomial is $f(x) = x^2 - 2x + 3$.We can rewrite this as $y = (x - 1)^2 + 2$,or $(x - 1)^2 = 1(y - 2)$.Comparing this with the standard form of a parabola $(x - h)^2 = 4a(y - k)$,we have $h = 1$,$k = 2$,and $4a = 1$,which means $a = 1/4$.The focus of the parabola $(x - h)^2 = 4a(y - k)$ is given by $(h, k + a)$.Substituting the values,the focus is $(1, 2 + 1/4) = (1, 9/4)$.

Similar Questions

If the coordinates of the ends of a focal chord of the parabola $x^2=4ay$ are $(x_1, y_1)$ and $(x_2, y_2)$,then

If the focus of a parabola is $(3, 0)$ and the length of its latus rectum is $8$,what is its vertex?

The equation of the parabola with $x+2y=1$ as directrix and $(1,0)$ as focus is

The condition for the line $y=mx+c$ to be a normal to the parabola $y^{2}=4ax$ is

If a normal is drawn to the parabola $y^2 = 4ax$ at the point $(2a, 2\sqrt{2}a)$,what is the length of the normal chord?

Vedclass Test Series

Exam Paper Generator

Online Exam Module