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

The length of the subnormal to any point of a curve is constant. Then,the eccentricity of the curve is . . . . . .

The point at which the line $y = mx + c$ touches the parabola $y^2 = 4ax$ is

The axis of a parabola is along the line $y=x$. The distance of its vertex $A$ from $(0,0)$ is $\sqrt{2}$ and that of its focus $S$ from $(0,0)$ is $2\sqrt{2}$. If $A$ and $S$ lie in the first quadrant,then the equation of the parabola in parametric form is

$A$ point $P(x, y)$ moves such that its distance from the point $(a, 0)$ is always equal to its distance from the line $x + a = 0$. The locus of the point is:

From an external point $P(h, k)$,a pair of tangent lines are drawn to the parabola $y^2 = 4x$. If $\theta_1$ and $\theta_2$ are the inclinations of these tangents with the $x$-axis such that $\theta_1 + \theta_2 = \frac{\pi}{4}$,then the locus of $P$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module