To solve for the points where the functions [tex]\( f_1(x) = 3x^2 - x + 2 \)[/tex] and [tex]\( f_2(x) = 2x - 5 \)[/tex] are equal, we need to set the functions equal to each other and solve for [tex]\( x \)[/tex].
1. Start by setting the two functions equal:
[tex]\[
3x^2 - x + 2 = 2x - 5
\][/tex]
2. Move all terms to one side of the equation to set it to zero:
[tex]\[
3x^2 - x + 2 - 2x + 5 = 0
\][/tex]
Simplify the equation:
[tex]\[
3x^2 - 3x + 7 = 0
\][/tex]
3. Now we have a quadratic equation:
[tex]\[
3x^2 - 3x + 7 = 0
\][/tex]
4. To solve this quadratic equation, use the quadratic formula, [tex]\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex], where [tex]\( a = 3 \)[/tex], [tex]\( b = -3 \)[/tex], and [tex]\( c = 7 \)[/tex].
5. Substitute the values of [tex]\( a \)[/tex], [tex]\( b \)[/tex], and [tex]\( c \)[/tex] into the quadratic formula:
[tex]\[
x = \frac{-(-3) \pm \sqrt{(-3)^2 - 4 \cdot 3 \cdot 7}}{2 \cdot 3}
\][/tex]
6. Simplify inside the square root:
[tex]\[
x = \frac{3 \pm \sqrt{9 - 84}}{6}
\][/tex]
[tex]\[
x = \frac{3 \pm \sqrt{-75}}{6}
\][/tex]
7. Since the discriminant (the value inside the square root) is negative, we have complex roots. Let's express the square root of [tex]\(-75\)[/tex]:
[tex]\[
\sqrt{-75} = \sqrt{75} \cdot i = 5\sqrt{3}i
\][/tex]
8. Substitute back into the quadratic formula:
[tex]\[
x = \frac{3 \pm 5\sqrt{3}i}{6}
\][/tex]
9. Simplify the fractions:
[tex]\[
x = \frac{3}{6} \pm \frac{5\sqrt{3}i}{6}
\][/tex]
[tex]\[
x = \frac{1}{2} \pm \frac{5\sqrt{3}i}{6}
\][/tex]
Therefore, the points where the functions [tex]\( f_1(x) \)[/tex] and [tex]\( f_2(x) \)[/tex] intersect are:
[tex]\[
x = \frac{1}{2} - \frac{5\sqrt{3}i}{6} \quad \text{and} \quad x = \frac{1}{2} + \frac{5\sqrt{3}i}{6}
\][/tex]
