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Problem 762

# Find, both analytically and graphically, the points of intersection of the two curves whose equations are $$2 x+y-4=0 \text { and } y^{2}-4 x=0$$

Expert verified
The points of intersection of the given curves are (1, 2) and (4, -4). This is determined analytically by solving the system of equations using the substitution method and then verifying graphically by plotting both equations on the same graph to observe their intersection points.
See the step by step solution

## Step 1: Write down the given equations

We are given the following equations: $$2x + y - 4 = 0 \quad (1) \\ y^2 - 4x = 0 \quad (2)$$

## Step 2: Solve the system of equations analytically

We will solve the system of equations using the substitution method. We can rewrite equation (1) in terms of y: $$y = 4 - 2x$$ Next, substitute this into equation (2) to solve for x: $$(4 - 2x)^2 - 4x = 0$$

## Step 3: Simplify and solve the quadratic equation for x

Simplify the equation obtained in the previous step: $$(16 - 16x + 4x^2) - 4x = 0 \\ 4x^2 - 20x + 16 = 0$$ Next, divide the equation by 4 to simplify it further: $$x^2 - 5x + 4 = 0$$ Now, factor the quadratic equation: $$(x - 4)(x - 1) = 0$$ From this, we find the two possible values of x: x = 1 and x = 4.

## Step 4: Find the corresponding y values

For each x value, find the corresponding y value by substituting back into the equation for y (from Step 2): $$y = 4 - 2x$$ For x = 1: $$y = 4 - 2(1) = 2$$ For x = 4: $$y = 4 - 2(4) = -4$$

## Step 5: Write down the intersection points

The points of intersection are the (x, y) pairs we found: $$(1, 2) \quad \text{and} \quad (4, -4)$$

## Step 6: Verify graphically

To verify graphically, we need to plot both equations on the same graph: 1. Plot the line of equation (1) with y = 4 - 2x. 2. Plot the parabola of equation (2) with y^2 = 4x. 3. Observe the points where the line and the parabola intersect. By doing this, you will find that the line and the parabola intersect at the points (1, 2) and (4, -4), which confirms our analytical solution.

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