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Magazine Chapter 11: Problem 39
Sketch the graph of the given equation. Find the intercepts; approximate to the nearest tenth where necessary. $$y=4 x^{2}-8 x-5$$
Short Answer
Expert verified
The intercepts are \((0, -5)\), \((2.5, 0)\), and \((-0.5, 0)\). The vertex is \((1, -9)\).
Step by step solution
01
- Identify the Type of Equation
Recognize that the given equation is a quadratic equation in the form of \[y = ax^2 + bx + c\]. In this case, \[a = 4\], \[b = -8\], and \[c = -5\].
02
- Find the Y-Intercept
To find the y-intercept, set \(x = 0\) and solve for \(y\). \[y = 4(0)^2 - 8(0) - 5 = -5\]. So, the y-intercept is \((0, -5)\).
03
- Find the X-Intercepts
To find the x-intercepts, set \(y = 0\) and solve for \(x\). \[0 = 4x^2 - 8x - 5\]. Use the quadratic formula \[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\].Substitute \(a = 4\), \(b = -8\), and \(c = -5\). \[x = \frac{8 \pm \sqrt{64 + 80}}{8}\], \[x = \frac{8 \pm \sqrt{144}}{8}\], \[x = \frac{8 \pm 12}{8}\]. So, \[x_1 = 2.5\] and \[x_2 = -0.5\]. The x-intercepts are \((2.5, 0)\) and \((-0.5, 0)\).
04
- Find the Vertex
The vertex of a parabola \(y = ax^2 + bx + c\) occurs at \(x = \frac{-b}{2a}\). Substitute \(a = 4\) and \(b = -8\). \[x = \frac{8}{8} = 1\]. Substitute \(x = 1\) back into the equation to find \(y\). \[y = 4(1)^2 - 8(1) - 5 = -9\]. So, the vertex is \((1, -9)\).
05
- Sketch the Graph
Plot the y-intercept \((0, -5)\), the x-intercepts \((2.5, 0)\) and \((-0.5, 0)\), and the vertex \((1, -9)\) on a coordinate plane. Draw a parabolic curve passing through these points, opening upwards as the coefficient \(4\) is positive.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Quadratic Equation
A quadratic equation is any equation that can be written in the form \ \ \(y = ax^2 + bx + c\). In the equation, \(a\), \(b\), and \(c\) are constants, with \(a eq 0\). This form is known as the standard form of a quadratic equation. Quadratic equations produce parabolas when graphed. The given equation, \(y = 4x^2 - 8x - 5\), is a prime example of a quadratic equation. Here, \(a = 4\), \(b = -8\), and \(c = -5\). The shape of the graph and its properties, like the direction of opening and position, depend on these coefficients.
X-Intercepts
X-intercepts are the points where the graph crosses the x-axis. At these points, \(y = 0\). To find the x-intercepts of a quadratic equation, you set the equation equal to zero and solve for \(x\).
For our equation, we solve \(0 = 4x^2 - 8x - 5\) using the quadratic formula: \ \ \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 4\), \(b = -8\), and \(c = -5\).
Substituting these values, we get \ \ \(x = \frac{8 \pm \sqrt{64 + 80}}{8}\), which simplifies to \ \ \(x = 2.5\) and \ \ \(x = -0.5\).
Hence, the x-intercepts are at the points \( (2.5, 0) \) and \( (-0.5, 0) \).
For our equation, we solve \(0 = 4x^2 - 8x - 5\) using the quadratic formula: \ \ \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 4\), \(b = -8\), and \(c = -5\).
Substituting these values, we get \ \ \(x = \frac{8 \pm \sqrt{64 + 80}}{8}\), which simplifies to \ \ \(x = 2.5\) and \ \ \(x = -0.5\).
Hence, the x-intercepts are at the points \( (2.5, 0) \) and \( (-0.5, 0) \).
Y-Intercept
The y-intercept is the point where the graph crosses the y-axis. At this point, \(x = 0\). To find the y-intercept, substitute \(x = 0\) into the quadratic equation.
For our equation: \ \ \(y = 4x^2 - 8x - 5\), substitute \(x = 0\): \ \ \(y = 4(0)^2 - 8(0) - 5\), so \(y = -5\).
Therefore, the y-intercept is the point \((0, -5)\).
For our equation: \ \ \(y = 4x^2 - 8x - 5\), substitute \(x = 0\): \ \ \(y = 4(0)^2 - 8(0) - 5\), so \(y = -5\).
Therefore, the y-intercept is the point \((0, -5)\).
Vertex
The vertex is a significant point of a parabola. It is the peak (if the parabola opens upwards) or the lowest point (if it opens downwards). For the quadratic equation \(y = ax^2 + bx + c\), the vertex occurs at \(x = \frac{-b}{2a}\).
In our equation, \(a = 4\) and \(b = -8\).
So, \ \ \(x = \frac{8}{8} = 1\).
Substitute \(x = 1\) back into the equation to find \(y\): \ \ \(y = 4(1)^2 - 8(1) - 5 = -9\).
Thus, the vertex is at the point \((1, -9)\).
The vertex helps determine the axis of symmetry for the parabola, which is the vertical line that passes through the vertex.
In our equation, \(a = 4\) and \(b = -8\).
So, \ \ \(x = \frac{8}{8} = 1\).
Substitute \(x = 1\) back into the equation to find \(y\): \ \ \(y = 4(1)^2 - 8(1) - 5 = -9\).
Thus, the vertex is at the point \((1, -9)\).
The vertex helps determine the axis of symmetry for the parabola, which is the vertical line that passes through the vertex.
Parabola
A parabola is the graph of a quadratic equation. It is a symmetrical curve that opens either upwards or downwards. The direction in which the parabola opens is determined by the coefficient \(a\):
Key features of a parabola include:
- If \(a > 0\), the parabola opens upwards.
- If \(a < 0\), it opens downwards.
Key features of a parabola include:
- The vertex: the highest or lowest point.
- The axis of symmetry: a vertical line through the vertex, dividing the parabola into two mirror images.
- The x-intercepts: points where the parabola crosses the x-axis.
- The y-intercept: the point where the parabola crosses the y-axis.
