find the vertex and y - intercept of the quadratic function y = 3x^2 - 12x + 40

Explanation:

Step1: Find x - coordinate of vertex

For a quadratic function $y = ax^{2}+bx + c$, the x - coordinate of the vertex is given by $x=-\frac{b}{2a}$. Here, $a = 5$, $b=-12$, so $x =-\frac{-12}{2\times5}=\frac{12}{10}=\frac{6}{5}$.

Step2: Find y - coordinate of vertex

Substitute $x = \frac{6}{5}$ into the function $y = 5x^{2}-12x + 40$. $y=5\times(\frac{6}{5})^{2}-12\times\frac{6}{5}+40$ $y = 5\times\frac{36}{25}-\frac{72}{5}+40$ $y=\frac{36}{5}-\frac{72}{5}+40$ $y=\frac{36 - 72}{5}+40$ $y=-\frac{36}{5}+40=\frac{-36 + 200}{5}=\frac{164}{5}$.

Step3: Find y - intercept

To find the y - intercept, set $x = 0$ in the function. When $x = 0$, $y=5\times0^{2}-12\times0 + 40=40$.

Answer:

Vertex: $(\frac{6}{5},\frac{164}{5})$; y - intercept: 40