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Chapter 2: Problem 7

Find the standard equation of any parabola that has vertex \(V\) $$f(x)=2 x^{2}-16 x+35$$

Short Answer

Expert verified
The vertex form of the given quadratic is \( f(x) = 2(x-4)^2 + 3 \).

Step by step solution

01

Identify the Standard Form

The standard form of a quadratic function is \(f(x) = ax^2 + bx + c\). Our given function, \(f(x) = 2x^2 - 16x + 35\), is already in this form with \(a = 2\), \(b = -16\), and \(c = 35\).
02

Find the Vertex

The vertex \((h, k)\) of a parabola in standard form \(f(x) = ax^2 + bx + c\) can be found using the formula \(h = -\frac{b}{2a}\). Substitute \(b = -16\) and \(a = 2\) into the formula to find \(h\).\[ h = -\frac{-16}{2 \times 2} = \frac{16}{4} = 4\]To find \(k\), substitute \(h\) back into the function:\[k = f(4) = 2(4)^2 - 16(4) + 35 = 2(16) - 64 + 35 = 32 - 64 + 35 = 3\]Thus, the vertex is \((4, 3)\).
03

Write the Vertex Form Equation

The vertex form of a parabola is given by \(f(x) = a(x-h)^2 + k\). We already have \(a = 2\), \(h = 4\), and \(k = 3\). Substitute these values into the vertex form equation: \[ f(x) = 2(x-4)^2 + 3 \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Standard Form of a Parabola
When working with quadratic functions, one of the key representations is the **standard form**. This is expressed as \[ f(x) = ax^2 + bx + c \] where
  • \(a\) is the coefficient of the quadratic term, indicating the parabola's direction and degree of stretch.
  • \(b\) is the coefficient of the linear term, affecting the placement of the axis of symmetry.
  • \(c\) is the constant term, which impacts where the parabola intersects the y-axis.
Understanding the standard form is crucial, as it provides immediate details about the parabola's shape and position. For example, if \(a > 0\), the parabola opens upwards, like a cup. If \(a < 0\), it opens downwards, like a frown. Additionally, changing \(b\) and \(c\) shifts the position of the parabola in the coordinate plane.
Vertex of a Parabola
The **vertex** of a parabola is a key feature, representing the highest or lowest point based on its orientation. To find the vertex of a parabola in standard form \( f(x) = ax^2 + bx + c \), you can use the formula for the vertex's \(h\)-coordinate:\[ h = -\frac{b}{2a} \]Once \(h\) is calculated, substitute it back into the original equation to find \(k\), which represents the vertex's \(y\)-coordinate. This gives the coordinates of the vertex as \((h, k)\).
  • Using the given equation \( f(x) = 2x^2 - 16x + 35 \), we computed \(h = 4\).
  • Substituting back, we found \(k = 3\).
  • Thus, the vertex of this particular parabola is \((4, 3)\).
Knowing the vertex helps in sketching the parabola accurately, as it gives you the turning point from which the parabola extends in both directions.
Vertex Form Equation
The **vertex form** of a quadratic function is a more visual representation given by:\[f(x) = a(x-h)^2 + k\]where
  • \(a\) determines the width and direction of the parabola just like in the standard form.
  • \((h, k)\) is the vertex, providing the exact position of the parabola's peak or trough.
Converting the standard form to the vertex form can be useful for easily identifying the vertex and understanding how the parabola is oriented in the graph. For the example \( f(x) = 2x^2 - 16x + 35 \), we derived that the vertex is \((4, 3)\) and substituted these into the vertex form to get:\[f(x) = 2(x-4)^2 + 3\]Using this form makes graphing more straightforward because it directly shows how the parabola is transformed from the parent function \(y = x^2\) by translation and scaling.

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Most popular questions from this chapter

(a) Use the quadratic formula to find the zeros of \(f .\) (b) Find the maximum or minimum value of \(f(x)\) (c) Sketch the graph of \(f\) $$f(x)=x^{2}+4 x+9$$

Graph \(y=x^{3}-x^{1 / 3}\) and \(f\) on the same \(c o-\) ordinate plane, and estimate the points of intersection. $$f(x)=x^{2}-x-\frac{1}{4}$$

If \(f(x)=\sqrt{x}-1\) and \(g(x)=x^{3}+1,\) approximate \((f \circ g)(0.0001) .\) In order to avoid calculating a zero value for \((f \circ g)(0.0001),\) rewrite the formula for \(f \circ g\) as $$\frac{x^{3}}{\sqrt{x^{3}+1}+1}$$

Solve the equation \((f \circ g)(x)=0\). $$f(x)=x^{2}-x-2, \quad g(x)=2 x-5$$

Find the standard equation of a parabola that has a vertical axis and satisfies the given conditions. Vertex \((4,-7), x\) -intercept \(-4\)
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