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Magazine Chapter 9: Problem 63
\(y=x^{2}-10 x+29\)
Short Answer
Expert verified
The vertex is (5, 4) and the parabola opens upwards. It does not have any real x-intercepts and has a y-intercept at 29.
Step by step solution
01
- Identify the quadratic function
Given the quadratic function: y = x^2 - 10x + 29
02
- Write the function in standard form
The given function is already in the standard form of a quadratic equation: y = ax^2 + bx + c where a = 1, b = -10, and c = 29.
03
- Find the vertex using the vertex formula
The vertex of a quadratic function can be found using the formula: \(x = \frac{-b}{2a}\)Substitute a and b: \[x = \frac{-(-10)}{2(1)} = \frac{10}{2} = 5\]So, the x-coordinate of the vertex is 5.
04
- Calculate the y-coordinate of the vertex
Substitute x = 5 back into the original equation to find y: \[y = 5^2 - 10(5) + 29 = 25 - 50 + 29 = 4\]So, the vertex is at (5, 4).
05
- Determine the direction of the parabola
Since the coefficient of \(x^2\) is positive (a = 1), the parabola opens upwards.
06
- Find the y-intercept
The y-intercept occurs when x = 0. Substitute x = 0 into the equation:\(y = 0^2 - 10(0) + 29 = 29\)So, the y-intercept is 29.
07
- Find the x-intercepts
Set the equation to 0 and solve for x: \(0 = x^2 - 10x + 29\)Use the quadratic formula: \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)Substituting a = 1, b = -10, and c = 29: \[x = \frac{10 \pm \sqrt{(-10)^2 - 4(1)(29)}}{2(1)} = \frac{10 \pm \sqrt{100 - 116}}{2} = \frac{10 \pm \sqrt{-16}}{2}\]Since the discriminant is negative, there are no real roots, so x-intercepts do not exist.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Quadratic Functions
A quadratic function is a polynomial function of degree 2. It takes the form \(y = ax^2 + bx + c\) where a, b, and c are constants, and \(a eq 0\). This results in a parabola when graphed. Parabolas can open upwards or downwards depending on the sign of the coefficient a. The general properties of quadratic functions are:
The equation given in the exercise is: \(y = x^2 - 10x + 29\). Here, \(a = 1\), so the parabola opens upwards.
- They have a vertex, which is the highest or lowest point.
- They are symmetric around a vertical line called the axis of symmetry.
- If a > 0, the parabola opens upwards; if a < 0, it opens downwards.
The equation given in the exercise is: \(y = x^2 - 10x + 29\). Here, \(a = 1\), so the parabola opens upwards.
Vertex Calculation
The vertex of a quadratic function is a crucial point that indicates the maximum or minimum value of the function. You can find the vertex using the vertex formula, which provides the x-coordinate of the vertex: \(x = \frac{-b}{2a}\).
The x-coordinate of the vertex is 5. To find the y-coordinate, substitute x = 5 back into the equation: \(y = 5^2 - 10(5) + 29 = 25 - 50 + 29 = 4\). Therefore, the vertex of the function is (5, 4). The vertex tells us that the parabola reaches its minimum value of 4 when x is 5.
- For the given function, \(y = x^2 - 10x + 29\), \(a = 1\) and \(b = -10\).
- Using the formula: \(x = \frac{-(-10)}{2(1)} = \frac{10}{2} = 5\).
The x-coordinate of the vertex is 5. To find the y-coordinate, substitute x = 5 back into the equation: \(y = 5^2 - 10(5) + 29 = 25 - 50 + 29 = 4\). Therefore, the vertex of the function is (5, 4). The vertex tells us that the parabola reaches its minimum value of 4 when x is 5.
Discriminant Analysis
The discriminant of a quadratic function, given by \(b^2 - 4ac\) from the quadratic formula, helps us determine the number and type of roots (solutions) of the equation.
For the equation \(0 = x^2 - 10x + 29\), substituting \(a = 1\), \(b = -10\), and \(c = 29\) into the discriminant formula gives us: \((-10)^2 - 4(1)(29) = 100 - 116 = -16\). Since the discriminant is negative, there are no real x-intercepts for this quadratic equation.
- If the discriminant is positive (> 0), there are two distinct real roots.
- If the discriminant is zero (= 0), there is exactly one real root (the parabola touches the x-axis).
- If the discriminant is negative (< 0), there are no real roots; instead, there are two complex roots.
For the equation \(0 = x^2 - 10x + 29\), substituting \(a = 1\), \(b = -10\), and \(c = 29\) into the discriminant formula gives us: \((-10)^2 - 4(1)(29) = 100 - 116 = -16\). Since the discriminant is negative, there are no real x-intercepts for this quadratic equation.
Standard Form of a Quadratic Equation
A quadratic equation in standard form is written as: \(y = ax^2 + bx + c\), where \(a\), \(b\), and \(c\) are constants. This form is helpful because it immediately tells you:
The given function \(y = x^2 - 10x + 29\) is already in standard form with \(a = 1\), \(b = -10\), and \(c = 29\). This makes it easy to apply other concepts like vertex calculation and discriminant analysis.
The y-intercept here is found by setting \(x\) to 0: \(y = 1(0)^2 - 10(0) + 29 = 29\). Therefore, the y-intercept is 29.
- The direction the parabola opens (based on the sign of \(a\)).
- Initial value when \(x = 0\), showing the y-intercept.
The given function \(y = x^2 - 10x + 29\) is already in standard form with \(a = 1\), \(b = -10\), and \(c = 29\). This makes it easy to apply other concepts like vertex calculation and discriminant analysis.
The y-intercept here is found by setting \(x\) to 0: \(y = 1(0)^2 - 10(0) + 29 = 29\). Therefore, the y-intercept is 29.
