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

• A. Vertex: $(1,1)$, $x$-intercept: $(5,0)$; $y$-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 \(x = -\frac{b}{2a}\) where \(a=4\) and \(b=-8\). Substitute \(x\) back into the equation to find the \(y\)-coordinate of the vertex.
3. Calculate the \(y\)-intercept by evaluating the function at \(x=0\), which gives \(y = 4(0)^2 - 8(0) + 5\).
4. Find the \(x\)-intercepts by setting \(y=0\) and solving the quadratic equation \$4x^2 - 8x + 5 = 0$ using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\).
5. Plot the vertex and intercepts on the graph, then sketch the parabola opening upwards since \(a=4\) is positive.