what is the vertex of $g(x)=8x^{2}-48x + 65$?\n(-3, -7)\n(3, -7)\n(24, -7)\n(-24, -7)
what is the vertex of $g(x)=8x^{2}-48x + 65$?\n(-3, -7)\n(3, -7)\n(24, -7)\n(-24, -7)
Answer
Explanation:
Step1: Identify coefficients
For the quadratic function $g(x)=ax^{2}+bx + c$, here $a = 8$, $b=-48$, $c = 65$.
Step2: Find x - coordinate of vertex
The formula for the x - coordinate of the vertex of a quadratic function is $x=-\frac{b}{2a}$. Substitute $a = 8$ and $b=-48$ into the formula: $x=-\frac{-48}{2\times8}=\frac{48}{16}=3$.
Step3: Find y - coordinate of vertex
Substitute $x = 3$ into the function $g(x)=8x^{2}-48x + 65$. Then $g(3)=8\times3^{2}-48\times3 + 65=8\times9-144 + 65=72-144 + 65=-7$.
Answer:
B. $(3,-7)$