To find the maximum or minimum of a quadratic equation, we need to determine the vertex of the parabola, given by the equation [tex]\( ax^2 + bx + c \)[/tex]. The vertex [tex]\((h, k)\)[/tex] of a quadratic function [tex]\( y = ax^2 + bx + c \)[/tex] can be found using the formulas:
[tex]\[ h = -\frac{b}{2a} \][/tex]
[tex]\[ k = f(h) = a \left( -\frac{b}{2a} \right)^2 + b \left( -\frac{b}{2a} \right) + c \][/tex]
Let's evaluate each equation step-by-step.
### 1) [tex]\( 2x^2 + 3x - 1 \)[/tex]
- Identify [tex]\( a = 2 \)[/tex], [tex]\( b = 3 \)[/tex], and [tex]\( c = -1 \)[/tex].
- Find the x-coordinate of the vertex [tex]\( h \)[/tex]:
[tex]\[ h = -\frac{b}{2a} = -\frac{3}{2(2)} = -\frac{3}{4} = -0.75 \][/tex]
- Now, substitute [tex]\( h \)[/tex] back into the equation to find [tex]\( k \)[/tex]:
[tex]\[ k = 2 \left( -0.75 \right)^2 + 3 \left( -0.75 \right) - 1 \][/tex]
[tex]\[ k = 2(0.5625) + 3(-0.75) - 1 \][/tex]
[tex]\[ k = 1.125 - 2.25 - 1 \][/tex]
[tex]\[ k = -2.125 \][/tex]
The vertex is [tex]\((-0.75, -2.125)\)[/tex]. Since the coefficient [tex]\(a = 2\)[/tex] is positive, the parabola opens upwards and the vertex represents a minimum point.
### 2) [tex]\( -x^2 + 4x - 7 \)[/tex]
- Identify [tex]\( a = -1 \)[/tex], [tex]\( b = 4 \)[/tex], and [tex]\( c = -7 \)[/tex].
- Find the x-coordinate of the vertex [tex]\( h \)[/tex]:
[tex]\[ h = -\frac{b}{2a} = -\frac{4}{2(-1)} = \frac{4}{2} = 2 \][/tex]
- Now, substitute [tex]\( h \)[/tex] back into the equation to find [tex]\( k \)[/tex]:
[tex]\[ k = -\left(2\right)^2 + 4\left(2\right) - 7 \][/tex]
[tex]\[ k = -4 + 8 - 7 \][/tex]
[tex]\[ k = -3 \][/tex]
The vertex is [tex]\((2, -3)\)[/tex]. Since the coefficient [tex]\(a = -1\)[/tex] is negative, the parabola opens downwards and the vertex represents a maximum point.
### 3) [tex]\( \frac{3}{2} x^2 - x \)[/tex]
- Identify [tex]\( a = \frac{3}{2} \)[/tex], [tex]\( b = -1 \)[/tex], and [tex]\( c = 0 \)[/tex].
- Find the x-coordinate of the vertex [tex]\( h \)[/tex]:
[tex]\[ h = -\frac{b}{2a} = -\frac{-1}{2\left(\frac{3}{2}\right)} = \frac{1}{3} \][/tex]
- Now, substitute [tex]\( h \)[/tex] back into the equation to find [tex]\( k \)[/tex]:
[tex]\[ k = \frac{3}{2} \left( \frac{1}{3} \right)^2 - \left( \frac{1}{3} \right) \][/tex]
[tex]\[ k = \frac{3}{2} \left( \frac{1}{9} \right) - \left( \frac{1}{3} \right) \][/tex]
[tex]\[ k = \frac{3}{18} - \left( \frac{1}{3} \right) \][/tex]
[tex]\[ k = \frac{1}{6} - \frac{2}{6} \][/tex]
[tex]\[ k = -\frac{1}{6} \][/tex]
[tex]\[ k = -0.1667 \][/tex] (approximately)
The vertex is [tex]\((0.3333, -0.1667)\)[/tex]. Since the coefficient [tex]\( a = \frac{3}{2} \)[/tex] is positive, the parabola opens upwards and the vertex represents a minimum point.
### Summary
1. For [tex]\( 2x^2 + 3x - 1 \)[/tex], the vertex is [tex]\((-0.75, -2.125)\)[/tex], and it is a minimum point.
2. For [tex]\( -x^2 + 4x - 7 \)[/tex], the vertex is [tex]\((2, -3)\)[/tex], and it is a maximum point.
3. For [tex]\( \frac{3}{2} x^2 - x \)[/tex], the vertex is [tex]\((0.3333, -0.1667)\)[/tex], and it is a minimum point.
