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Magazine Chapter 11: Problem 18
Sketch the graph of the given equation. Find the intercepts; approximate to the nearest tenth where necessary. $$y=x^{2}-6 x+5$$
Short Answer
Expert verified
Y-intercept: (0,5), X-intercepts: (1,0) and (5,0), Vertex: (3,-4).
Step by step solution
01
Identify the equation type
The given equation is a quadratic equation in the form \(y = ax^2 + bx + c\). Here, \(a = 1\), \(b = -6\), and \(c = 5\).
02
Find the y-intercept
The y-intercept is the value of \(y\) when \(x = 0\). Substitute \(x = 0\) into the equation: \[ y = (0)^2 - 6(0) + 5 = 5 \] So, the y-intercept is \( (0, 5) \).
03
Find the x-intercepts
The x-intercepts are the values of \(x\) when \(y = 0\). Set \(y = 0\) and solve for \(x\): \[ 0 = x^2 - 6x + 5 \] Factor the quadratic equation: \[ 0 = (x - 1)(x - 5) \] Thus, the x-intercepts are at \(x = 1\) and \(x = 5\). So, the points are \( (1, 0) \) and \( (5, 0) \).
04
Find the vertex
The vertex of a parabola in the form \( y = ax^2 + bx + c \) can be found using \( x = -\frac{b}{2a} \). Substitute the values of \(a\) and \(b\): \[ x = -\frac{-6}{2(1)} = 3 \] Now, find the corresponding \(y\) value by substituting \(x = 3\) back into the equation: \[ y = 3^2 - 6(3) + 5 = -4 \] So, the vertex is at \( (3, -4) \).
05
Sketch the graph
Plot the points \( (0, 5) \) (y-intercept), \( (1, 0) \) (x-intercept), \( (5, 0) \) (x-intercept), and \( (3, -4) \) (vertex). Draw a smooth curve through these points to sketch the parabola.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
y-intercept
The y-intercept of a quadratic equation is the point where the graph crosses the y-axis. To find this point, you set the value of x to zero and solve for y in the equation.
For the equation given in the exercise, \( y = x^2 - 6x + 5 \), substitute x with 0.
\[ y = (0)^2 - 6(0) + 5 = 5 \]
So, the y-intercept is \text {(0, 5)}.
Identifying the y-intercept is crucial as it provides a starting point for sketching the graph.
For the equation given in the exercise, \( y = x^2 - 6x + 5 \), substitute x with 0.
\[ y = (0)^2 - 6(0) + 5 = 5 \]
So, the y-intercept is \text {(0, 5)}.
- Always remember: when x = 0, y = c in a quadratic equation \( y = ax^2 + bx + c \).
- This is a key feature that helps in graphing the equation.
Identifying the y-intercept is crucial as it provides a starting point for sketching the graph.
x-intercepts
The x-intercepts of a quadratic equation are the points where the graph crosses the x-axis. To find the x-intercepts, set the value of y to zero and solve for x. This involves factoring the quadratic equation, finding the roots, or using the quadratic formula.
For the exercise's equation \( y = x^2 - 6x + 5 \), set y to 0 and solve for x:
\[ 0 = x^2 - 6x + 5 \]
Factor the quadratic:
\[ 0 = (x - 1)(x - 5) \]
This means the x-intercepts are at \( x = 1 \) and \( x = 5 \). So the points are \text {(1, 0)} and \text {(5, 0)}.
Finding x-intercepts is important as they define the points where the parabola intersects the x-axis.
For the exercise's equation \( y = x^2 - 6x + 5 \), set y to 0 and solve for x:
\[ 0 = x^2 - 6x + 5 \]
Factor the quadratic:
\[ 0 = (x - 1)(x - 5) \]
This means the x-intercepts are at \( x = 1 \) and \( x = 5 \). So the points are \text {(1, 0)} and \text {(5, 0)}.
- Factoring quadratics can sometimes be challenging, but it's a useful technique for finding x-intercepts quickly.
- Alternatively, you can use the quadratic formula \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] if necessary.
Finding x-intercepts is important as they define the points where the parabola intersects the x-axis.
vertex of a parabola
The vertex of a parabola is the maximum or minimum point on the graph, depending on whether the parabola opens upwards or downwards. For a quadratic equation, the x-coordinate of the vertex can be found using \[ x = -\frac{b}{2a} \].
In the exercise's equation \( y = x^2 - 6x + 5 \), a = 1 and b = -6. So,
\[ x = -\frac{-6}{2(1)} = 3 \]
To find the corresponding y-coordinate, substitute x = 3 back into the equation:
\[ y = 3^2 - 6(3) + 5 = -4 \]
Thus, the vertex is \text {(3, -4)}.
Knowing the vertex helps you understand the shape and direction of the parabola.
In the exercise's equation \( y = x^2 - 6x + 5 \), a = 1 and b = -6. So,
\[ x = -\frac{-6}{2(1)} = 3 \]
To find the corresponding y-coordinate, substitute x = 3 back into the equation:
\[ y = 3^2 - 6(3) + 5 = -4 \]
Thus, the vertex is \text {(3, -4)}.
- The vertex is crucial because it represents the turning point of the parabola.
- In this example, the vertex indicates the lowest point on the graph since the parabola opens upwards (a > 0).
Knowing the vertex helps you understand the shape and direction of the parabola.
factoring quadratics
Factoring quadratics involves expressing the quadratic equation in the form \( (x - p)(x - q) \), where p and q are the roots of the equation. This method is used to find the x-intercepts quickly.
Consider the exercise's equation \( y = x^2 - 6x + 5 \). To factor it, look for two numbers that multiply to give the constant term (c) and add to give the coefficient of x (b).
Here, -1 and -5 multiply to 5 and add to -6, so:
\[ y = (x - 1)(x - 5) \]
This reveals the roots, p = 1 and q = 5.
Mastering factoring quadratics makes solving and graphing these equations simpler.
Consider the exercise's equation \( y = x^2 - 6x + 5 \). To factor it, look for two numbers that multiply to give the constant term (c) and add to give the coefficient of x (b).
Here, -1 and -5 multiply to 5 and add to -6, so:
\[ y = (x - 1)(x - 5) \]
This reveals the roots, p = 1 and q = 5.
- Factoring is efficient and often more intuitive than using the quadratic formula.
- For more complex quadratics, use the quadratic formula or complete the square.
Mastering factoring quadratics makes solving and graphing these equations simpler.
