Questions: Quadratic Functions Question 6, 2.1.57 Part 5 of 6 For the equation y=x^2-2x+6, complete parts a) through e). (a) Which way does the parabola open? A. The parabola opens upward because a is positive. B. The parabola opens downward because a is negative. C. The parabola opens upward because a is negative. D. The parabola opens downward because a is positive. b) Find the vertex of the parabola. The vertex is (1,5). (Type an ordered pair.) c) Find the axis of symmetry. The axis of symmetry is x=1. (Type an equation.) (d) Identify the x-intercept(s). Select the correct choice below and, if necessary, fill in the answer box to complete your choice. A. The x-intercept(s) is/are x= . (Type an integer or a fraction. Use a comma to separate answers as needed.) B. There are no x-intercepts. Identify the y-intercept(s). Select the correct choice below and, if necessary, fill in the answer box to complete your choice. A. The y-intercept(s) is/are y= . (Type an integer or a fraction. Use a comma to separate answers as needed.) B. There are no y-intercepts.

Solution Steps

(a) To determine the direction in which the parabola opens, we need to look at the coefficient of the \(x^2\) term. If the coefficient (a) is positive, the parabola opens upward. If it is negative, the parabola opens downward.

(b) The vertex of a parabola given by \(y = ax^2 + bx + c\) can be found using the formula \((h, k)\), where \(h = -b/(2a)\) and \(k = f(h)\).

(c) The axis of symmetry for a parabola given by \(y = ax^2 + bx + c\) is the vertical line \(x = h\), where \(h\) is the x-coordinate of the vertex.

(d) To find the x-intercepts, we need to solve the equation \(x^2 - 2x + 6 = 0\). If the discriminant (\(b^2 - 4ac\)) is negative, there are no real x-intercepts. If it is zero or positive, we can find the x-intercepts using the quadratic formula.

(e) To find the y-intercept, we set \(x = 0\) in the equation and solve for \(y\).

The given quadratic equation is \( y = x^2 - 2x + 6 \). The coefficient of \( x^2 \) is \( a = 1 \), which is positive. Therefore, the parabola opens upward.

The vertex of a parabola given by \( y = ax^2 + bx + c \) can be found using the formula \( (h, k) \), where \( h = -\frac{b}{2a} \) and \( k = f(h) \).

Given: \[ a = 1, \quad b = -2, \quad c = 6 \]

Calculate \( h \): \[ h = -\frac{-2}{2 \cdot 1} = 1 \]

Calculate \( k \): \[ k = 1^2 - 2 \cdot 1 + 6 = 5 \]

Thus, the vertex is \( (1, 5) \).

The axis of symmetry for the parabola is the vertical line \( x = h \).

Given \( h = 1 \), the axis of symmetry is: \[ x = 1 \]

To find the x-intercepts, solve the equation \( x^2 - 2x + 6 = 0 \).

Calculate the discriminant: \[ \Delta = b^2 - 4ac = (-2)^2 - 4 \cdot 1 \cdot 6 = 4 - 24 = -20 \]

Since the discriminant is negative (\( \Delta < 0 \)), there are no real x-intercepts.

To find the y-intercept, set \( x = 0 \) in the equation \( y = x^2 - 2x + 6 \).

Calculate \( y \): \[ y = 0^2 - 2 \cdot 0 + 6 = 6 \]

Thus, the y-intercept is \( y = 6 \).

Final Answer