The minimum value of the expression 2x 2+ 5x + 5 is
15/8
Understanding the Minimum Value of a Quadratic Expression
The question asks for the minimum value of the expression \(2x^2 + 5x + 5\). This is a quadratic expression in the standard form \(ax^2 + bx + c\), where:
- \(a = 2\)
- \(b = 5\)
- \(c = 5\)
For a quadratic expression \(ax^2 + bx + c\), the graph is a parabola. If the coefficient of the \(x^2\) term, \(a\), is positive (which is the case here, as \(a=2 > 0\)), the parabola opens upwards. An upward-opening parabola has a minimum point at its vertex.
Finding the Vertex of the Parabola
The vertex of the parabola \(y = ax^2 + bx + c\) has an x-coordinate given by the formula:
\(x = -\frac{b}{2a}\)
Substituting the values from our expression:
\(x = -\frac{5}{2 \times 2}\)
\(x = -\frac{5}{4}\)
The minimum value of the expression occurs at this x-coordinate. To find the minimum value, we substitute this value of \(x\) back into the original expression \(2x^2 + 5x + 5\).
Calculating the Minimum Value
Substitute \(x = -\frac{5}{4}\) into \(2x^2 + 5x + 5\):
\(2\left(-\frac{5}{4}\right)^2 + 5\left(-\frac{5}{4}\right) + 5\)
First, calculate the square term:
\(\left(-\frac{5}{4}\right)^2 = \frac{(-5)^2}{(4)^2} = \frac{25}{16}\)
Now substitute this back:
\(2\left(\frac{25}{16}\right) + 5\left(-\frac{5}{4}\right) + 5\)
Perform the multiplications:
\(\frac{2 \times 25}{16} - \frac{5 \times 5}{4} + 5\)
\(\frac{50}{16} - \frac{25}{4} + 5\)
Simplify the first term and find a common denominator (16) for all terms:
\(\frac{25}{8} - \frac{25 \times 4}{4 \times 4} + \frac{5 \times 16}{16}\)
\(\frac{25}{8} - \frac{100}{16} + \frac{80}{16}\)
Convert all terms to have a denominator of 8 for easier calculation, or use 16:
Using denominator 8:
\(\frac{25}{8} - \frac{50}{8} + \frac{40}{8}\)
Now combine the numerators:
\(\frac{25 - 50 + 40}{8}\)
\(\frac{-25 + 40}{8}\)
\(\frac{15}{8}\)
Alternatively, the minimum value of \(ax^2 + bx + c\) for \(a>0\) can be directly calculated using the formula for the y-coordinate of the vertex:
\(y_{min} = c - \frac{b^2}{4a}\)
Substitute \(a=2\), \(b=5\), \(c=5\):
\(y_{min} = 5 - \frac{5^2}{4 \times 2}\)
\(y_{min} = 5 - \frac{25}{8}\)
To subtract, find a common denominator:
\(y_{min} = \frac{5 \times 8}{8} - \frac{25}{8}\)
\(y_{min} = \frac{40}{8} - \frac{25}{8}\)
\(y_{min} = \frac{40 - 25}{8}\)
\(y_{min} = \frac{15}{8}\)
Summary of Calculation
| Expression | Form \(ax^2+bx+c\) | Coefficients | Vertex x-coord (\(-b/2a\)) | Minimum Value Formula (\(c - b^2/4a\)) | Result |
|---|---|---|---|---|---|
| \(2x^2+5x+5\) | Yes | \(a=2, b=5, c=5\) | \(-\frac{5}{4}\) | \(5 - \frac{5^2}{4 \times 2} = 5 - \frac{25}{8}\) | \(\frac{15}{8}\) |
Both methods confirm that the minimum value of the expression \(2x^2 + 5x + 5\) is \(\frac{15}{8}\).
Revision Table: Quadratic Expressions
| Concept | Description | Formula/Property |
|---|---|---|
| Quadratic Expression | An expression of the form \(ax^2 + bx + c\) where \(a \neq 0\). | \(ax^2 + bx + c\) |
| Parabola Shape | Determined by the sign of \(a\). Opens up if \(a > 0\), opens down if \(a < 0\). | \(a > 0\) → Minimum value at vertex \(a < 0\) → Maximum value at vertex |
| Vertex x-coordinate | The x-value where the minimum (or maximum) occurs. | \(x_v = -\frac{b}{2a}\) |
| Vertex y-coordinate (Min/Max Value) | The minimum (or maximum) value of the expression. | \(y_v = c - \frac{b^2}{4a}\) or substitute \(x_v\) into \(ax^2 + bx + c\) |
Additional Information on Finding Minimum/Maximum
Understanding how to find the minimum or maximum value of a quadratic expression is a fundamental concept in algebra and calculus. Here's a bit more context:
- Completing the Square: Another method to find the minimum or maximum value is by completing the square. The expression \(ax^2 + bx + c\) can be rewritten in the form \(a(x-h)^2 + k\), where \((h, k)\) is the vertex. For \(a>0\), the term \(a(x-h)^2\) is always non-negative, so the minimum value occurs when \(a(x-h)^2 = 0\), i.e., at \(x=h\), and the minimum value is \(k\). For our expression \(2x^2 + 5x + 5\), completing the square would give: \(2(x^2 + \frac{5}{2}x) + 5\) \(2(x^2 + \frac{5}{2}x + (\frac{5}{4})^2 - (\frac{5}{4})^2) + 5\) \(2((x + \frac{5}{4})^2 - \frac{25}{16}) + 5\) \(2(x + \frac{5}{4})^2 - \frac{50}{16} + 5\) \(2(x + \frac{5}{4})^2 - \frac{25}{8} + \frac{40}{8}\) \(2(x + \frac{5}{4})^2 + \frac{15}{8}\) The minimum value is \(\frac{15}{8}\) at \(x = -\frac{5}{4}\). This method is more involved but provides the vertex coordinates directly.
- Calculus Method: Using calculus, the minimum or maximum of a function \(f(x)\) occurs where its derivative \(f'(x)\) is zero. For \(f(x) = 2x^2 + 5x + 5\), the derivative is \(f'(x) = 4x + 5\). Setting \(f'(x) = 0\) gives \(4x + 5 = 0\), so \(x = -\frac{5}{4}\). This confirms the x-coordinate of the vertex. Substituting this back into the original function gives the minimum value.
All these methods yield the same result, confirming the minimum value of the expression \(2x^2 + 5x + 5\) is \(\frac{15}{8}\).
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