Let's start with each problem step by step.
Problem 2: Completing the Square
We need to complete the square for the equation \( y = -x^2 + 8x + 7 \).
-
Rearranging the equation:
First, we can factor out \(-1\) from the terms involving \(x\):\[ y = - (x^2 - 8x) + 7 \]
-
Completing the square:
Next, we need to complete the square for \(x^2 - 8x\). Take half of the coefficient of \(x\) (which is \(-8\)), square it, and add/subtract it inside the bracket. Half of \(-8\) is \(-4\), and squaring it gives \(16\):\[ y = - (x^2 - 8x + 16 - 16) + 7 \]
This simplifies to:
\[ y = - ((x - 4)^2 - 16) + 7 \] \[ y = - (x - 4)^2 + 16 + 7 \] \[ y = - (x - 4)^2 + 23 \]
-
Identifying the vertex:
The vertex form of a parabola is \(y = a(x - h)^2 + k\), where \((h, k)\) is the vertex. From the completed square, we see that:- \(h = 4\)
- \(k = 23\)
Thus, the vertex is \((4, 23)\).
-
Finding the axis of symmetry:
The axis of symmetry is the vertical line that passes through the vertex. It is given by:\[ x = h \]
Therefore, the axis of symmetry is:
\[ x = 4 \]
Summary for Question 2:
- Vertex: \((4, 23)\)
- Axis of Symmetry: \(x = 4\)
Problem 3: Finding the Zeros of the Quadratic Equation
We have the quadratic equation \(y = 6x^2 + 13x + 5\). We will use the quadratic formula to find the zeros of the equation.
-
Identifying coefficients:
The quadratic formula is given by\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \(a = 6\), \(b = 13\), and \(c = 5\).
-
Calculating the discriminant:
First, we calculate the discriminant (\(D\)):\[ D = b^2 - 4ac = 13^2 - 4(6)(5) \] \[ D = 169 - 120 = 49 \]
-
Applying the quadratic formula:
\[ x = \frac{-13 \pm \sqrt{49}}{2(6)} = \frac{-13 \pm 7}{12} \]
Now we have two cases:
- Case 1: \(x = \frac{-13 + 7}{12} = \frac{-6}{12} = -\frac{1}{2}\)
- Case 2: \(x = \frac{-13 - 7}{12} = \frac{-20}{12} = -\frac{5}{3}\)
Summary for Question 3:
The zeros of the quadratic equation \(y = 6x^2 + 13x + 5\) are:
- \(x = -\frac{1}{2}\)
- \(x = -\frac{5}{3}\)
1 answer