CRE 5 - Graph & Maximum or Minimum value of Quadratic function | Algebra - Quadratic Equations
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Find the minimum value of the expression: x2 - 4x + 10.
- (a)
10
- (b)
5
- (c)
-6
- (d)
6
- (e)
None of these
Answer: Option D
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Explanation :
We know, that minimum value of ax2 + bx + c occurs at x = -b/2a
∴ Minimum value of x2 - 4x + 10 occurs at x = -(-4)/2 × 1 = 2.
∴ The minimum value is (2)2 - 4 × 2 + 10 = 4 - 8 + 10 = 6
Hence, option (d).
Workspace:
Find the maximum value of the expression: -2x2 - 4x + 11.
- (a)
12
- (b)
11
- (c)
13
- (d)
14
- (e)
None of these
Answer: Option C
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Explanation :
We know, that maximum value of ax2 + bx + c occurs at x = -b/2a
∴ Maximum value of -2x2 - 4x + 11 occurs at x = -(-4)/2 × -2 = -1.
∴ The minimum value is -2 × (-1)2 - 4 × (-1) + 11 = -2 + 4 + 11 = 13
Hence, option (c).
Workspace:
f(x) is a quadratic expression such that the coefficient of x2 is negative. If the roots of f(x) = 0 lie in the interval (-1, 1), then which of the following is necessarily true?
- (a)
f(2) > 0 & f(2) > 0
- (b)
f(2) > 0 & f(-2) < 0
- (c)
f(2) < 0 & f(-2) > 0
- (d)
f(2) < 0 & f(-2) < 0
- (e)
Cannot be determined
Answer: Option D
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Explanation :
Since the coefficient of x2 is negative, the graph of the quadratic function will be downward facing.
Also, since roots lie between -1 and 1, it means the graph will cut the x-axis between x = -1 and x = 1.
Hence, the graph will look something like this.
From the graph it is clear that the value of graph at x = -2 and x = 2 is negative.
Hence, f(-2) and f(2) both will be negative.
Hence, option (d).
Workspace:
f(x) is a quadratic expression such that the coefficient of x2 is positive. If the roots of f(x) = 0 lie on either sides of x = -1, then which of the following is necessarily true about f(-1)?
- (a)
f(-1) > 0
- (b)
f(-1) < 0
- (c)
f(-1) = 0
- (d)
Cannot be determined
Answer: Option B
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Explanation :
Since the coefficient of x2 is positive, the graph of the quadratic function will be upward facing such that its roots lie on either sides of x = -1.
The graph would look something like this.
From the graph it is clear that f(-1) is less than zero.
Hence, option (b).
Workspace:
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