The graph of p(x) = x^{2} + 6x + 9 intersects | vedclass

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

Question

(C) To determine how the graph of $p(x) = x^{2} + 6x + 9$ intersects the $X$-axis,we find the zeros of the polynomial by setting $p(x) = 0$.$x^{2} + 6

Vedclass · Accepted Answer

touches

Vedclass · Answer

(C) To determine how the graph of $p(x) = x^{2} + 6x + 9$ intersects the $X$-axis,we find the zeros of the polynomial by setting $p(x) = 0$.$x^{2} + 6x + 9 = 0$This is a perfect square trinomial,which can be factored as $(x + 3)^{2} = 0$.Thus,$x = -3$ is the only root (a repeated root).Since there is only one distinct real root,the parabola represented by the quadratic polynomial touches the $X$-axis at exactly one point,which is $(-3, 0)$.

The graph of $p(x) = x^{2} + 6x + 9$ intersects the $X$-axis $\ldots \ldots \ldots \ldots$

Similar Questions

Find the division of the following by synthetic division method: $p(x) = x^{3} - 4x^{2} + 5x + 3$ by $x - 2$.

Draw the graph of $p(x) = x^3 - 2x^2$ and find the zeros of this polynomial.

For the given sum and product of zeroes,find a quadratic polynomial. Also,find the zeroes of this polynomial by factorisation:
Sum of zeroes = $\frac{-3}{2 \sqrt{5}}$,Product of zeroes = $-\frac{1}{2}$

Find the zeroes of the following polynomial by the factorisation method and verify the relationship between the zeroes and the coefficients of the polynomial:
$7 y^{2}-\frac{11}{3} y-\frac{2}{3}$

The zeros of a polynomial $p(x) = x^{2} - 2x - 15$ are........

Vedclass Test Series

Exam Paper Generator

Online Exam Module