Question

identify the graph of $f(x)=x^{2}-8x + 15$.

Explanation:

Step1: Find the x - intercepts

Set $f(x)=x^{2}-8x + 15=0$. Factor the quadratic equation: $(x - 3)(x - 5)=0$. Then $x=3$ or $x = 5$.

Step2: Find the vertex

The x - coordinate of the vertex of a quadratic function $y=ax^{2}+bx + c$ is $x=-\frac{b}{2a}$. For $y=x^{2}-8x + 15$, where $a = 1$, $b=-8$, $c = 15$, $x=-\frac{-8}{2\times1}=4$. Substitute $x = 4$ into the function: $y=4^{2}-8\times4 + 15=16-32 + 15=-1$. The vertex is $(4,-1)$.

Step3: Analyze the parabola

Since $a = 1>0$, the parabola opens upwards. The graph has x - intercepts at $x = 3$ and $x = 5$ and vertex at $(4,-1)$.

Answer:

The graph with x - intercepts at $x = 3$ and $x = 5$ and vertex at $(4,-1)$ and opening upwards.