Solving a quadratic equation, it is found that:
For the first point of intersection:
- The x-coordinate of the point of intersection is 4.
- The y-coordinate of the point of intersection is 3.
For the second point of intersection:
- The x-coordinate of the point of intersection is -0.5.
- The y-coordinate of the point of intersection is 5.25.
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The equations are:
[tex]y_1 = x^2 - 4x + 3[/tex]
[tex]y_2 = -0.5x + 5[/tex]
The x-value at which they intersect is found considering:
[tex]y_1 = y_2[/tex]
Thus
[tex]x^2 - 4x + 3 = -0.5x + 5[/tex]
[tex]x^2 - 3.5x - 2 = 0[/tex]
Which is a quadratic equation with [tex]a = 1, b = -3.5, c = -2[/tex]
Applying Bhaskara, we get that:
[tex]\Delta = b^2 - 4ac = (-3.5)^2 - 4(1)(-2) = 20.25[/tex]
[tex]x_{1} = \frac{-(-3.5) + \sqrt{20.25}}{2} = 4[/tex]
[tex]x_{2} = \frac{-(-3.5) - \sqrt{20.25}}{2} = -0.5[/tex]
When x = 4, [tex]y = -0.5(4) + 5 = 3[/tex]
When x = -0.5, [tex]y = -0.5(-0.5) + 5 = 5.25[/tex]
Thus, the intersection points are (4,3) and (-0.5, 5.25).
A similar problem is given at https://brainly.com/question/16747531
