Graph the following quadratics and state its vertex and intercepts.
$y=4x^2-8x+5$

• A. Vertex: $\(\left\)(1,1\(\right\))$, $x$-intercept: $\(\left\)(5,0\(\right\))$; $y$-intercept: None
• B. Vertex: $\(\left\)(1,1\(\right\))$, $x$-intercept: None; $y$-intercept: $\(\left\)(0,5\(\right\))$
• C. Vertex: $\(\left\)(1,1\(\right\))$, $x$-intercept: $\(\left\)(5,0\(\right\))$; $y$-intercept: None
• D. Vertex: $\(\left\)(1,1\(\right\))$, $x$-intercept: None; $y$-intercept: $\(\left\)(0,5\(\right\))$

Answer: (B)

Verified step by step guidance

1. Identify the quadratic equation given: \(y = 4x^2 - 8x + 5\).
2. Find the vertex using the vertex formula for a quadratic \(y = ax^2 + bx + c\), where the x-coordinate of the vertex is \(x = -\frac{b}{2a}\). Substitute \(a=4\) and \(b=-8\) to find \(x\).
3. Calculate the y-coordinate of the vertex by substituting the x-value found into the original equation \(y = 4x^2 - 8x + 5\).
4. Find the y-intercept by evaluating the quadratic at \(x=0\), which is simply the constant term \(c\) in the equation.
5. Find the x-intercepts by setting \(y=0\) and solving the quadratic equation \$4x^2 - 8x + 5 = 0$ using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\).