Graph the following quadratics and state its vertex and intercepts.
$y=-3x^2+10x-3$

• A. Vertex: $\(\left\)(-\(\frac\)53,\(\frac{16}{3}\]\right\))$; $x$-intercept: $\(\left\)(-\(\frac\)13,0\(\right\)),\(\left\)(-3,0\(\right\))$; $y$-intercept: $\(\left\)(0,-3\(\right\))$
• B. Vertex: $\(\left\)(-\(\frac\)53,\(\frac{16}{3}\]\right\))$; $x$-intercept: $\(\left\)(-\(\frac\)13,0\(\right\)),\(\left\)(-3,0\(\right\))$; $y$-intercept: $\(\left\)(0,-3\(\right\))$
• C. Vertex: $\(\left\)(\(\frac\)53,\(\frac{16}{3}\]\right\))$; $x$-intercept: $\(\left\)(\(\frac\)13,0\(\right\)),\(\left\)(-3,0\(\right\))$; $y$-intercept: $\(\left\)(0,-3\(\right\))$
• D. Vertex: $\(\left\)(\(\frac\)53,\(\frac{16}{3}\]\right\))$; $x$-intercept: $\(\left\)(\(\frac\)13,0\(\right\)),\(\left\)(-3,0\(\right\))$; $y$-intercept: $\(\left\)(0,-3\(\right\))$

Answer: (D)

Verified step by step guidance

1. Identify the quadratic function given: \(y = -3x^2 + 10x - 3\).
2. Find the vertex using the vertex formula for a parabola \(y = ax^2 + bx + c\), where the x-coordinate of the vertex is \(x = -\frac{b}{2a}\). Substitute \(a = -3\) and \(b = 10\) to find \(x\).
3. Calculate the y-coordinate of the vertex by substituting the x-value found into the original quadratic equation.
4. Find the x-intercepts by setting \(y = 0\) and solving the quadratic equation \(-3x^2 + 10x - 3 = 0\) using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\).
5. Find the y-intercept by evaluating the quadratic function at \(x = 0\), which gives \(y = c\).