Question
The curves of $y = 2x^2$ and $y = 4x$ intersect each other at
The correct answer is
exactly two points.
Finding Intersection Points of $y = 2x^2$ and $y = 4x$
To find where the curves $y = 2x^2$ and $y = 4x$ intersect, we set the equations equal to each other.
Solving the System of Equations
-
Set the expressions for $y$ equal:
$2x^2 = 4x$
-
Rearrange the equation to form a quadratic equation:
$2x^2 - 4x = 0$
-
Factor the equation:
$2x(x - 2) = 0$
-
Solve for $x$ by setting each factor to zero:
- $2x = 0 \implies x = 0$
- $x - 2 = 0 \implies x = 2$
-
Find the corresponding $y$ values using $y = 4x$:
- For $x = 0$, $y = 4(0) = 0$. The point is $(0, 0)$.
- For $x = 2$, $y = 4(2) = 8$. The point is $(2, 8)$.
Conclusion on Intersection Points
The calculations show two distinct solutions for $x$, leading to two distinct intersection points, $(0, 0)$ and $(2, 8)$. Therefore, the curves intersect at exactly two points.
Was this answer helpful?
Important Questions from Algebra (Notes)
- What is the remainder when 2023²⁰²⁴ + 2025²⁰²⁴ is divided by 2024?
In an examination, a student scores 4 marks for every correct answer and loses 1 mark for every wrong answer. If she/he attempts all 60 questions and secures 130 marks, the number of questions she/he attempts wrongly, are?
Match List-I with List-II
List-1 List-II (A) If $\begin{bmatrix}\lambda-1 & 0 \\ 0 & \lambda-1 \end{bmatrix} $, then $\lambda$ is (I) 0 (B) If A=$ \begin{bmatrix}1 & 2 \\2 & 4 \end{bmatrix} $, then $\Delta$ is (II) 1 (C) If A = $ \begin{bmatrix}1 & 0 \\0 & \frac{1}{2} \end{bmatrix} $, then $|A^{-1}|$ is (III) -2 (D) If $ \begin{bmatrix}a+1 & 1 \\1 & 2 \end{bmatrix} = \begin{bmatrix}-1 & 1 \\1 & 2 \end{bmatrix} $, then a is (IV) 2 Choose the correct answer from the options given below:
- If (x - 1) is a factor of $2x^2 - 5x + k = 0$, then the value of k is:
- If $x = (2+\sqrt{3})^{\frac{1}{3}} + (2+\sqrt{3})^{-\frac{1}{3}}$ and $x^3-3x + k = 0$, then the value of k is:
Need Expert Advice?
