Question

1. use quadratic equation in factored form: f(x)=(x - 3)(x - 1)

a) what is the standard form of given equation?
f(x)=x² - 4x + 3
b) what are the coordinates of the y - intercept? (0,3)
c) what are the coordinates of each of the x - intercepts? set each given factor to zero. put answers here: 1) (1,0) 2) (3,0)
d) what is the x - coordinate of the vertex? 2
show work here: x=-b/2a=-(-4)/(2×1)=4/2 = 2
e) what is the y - coordinate of the vertex?
show work here:

Explanation:

Step1: Recall standard - form formula

For a quadratic function $f(x)=(x - p)(x - q)$, expand using FOIL. Here $p = 3$ and $q = 1$, so $f(x)=x^{2}-x-3x + 3=x^{2}-4x + 3$.

Step2: Find y - intercept

The y - intercept is found by setting $x = 0$. Substitute $x = 0$ into $f(x)=x^{2}-4x + 3$, then $f(0)=0^{2}-4\times0 + 3=3$. So the coordinates of the y - intercept are $(0,3)$.

Step3: Find x - intercepts

Set $f(x)=0$, so $(x - 3)(x - 1)=0$. By the zero - product property, $x-3 = 0$ gives $x = 3$ and $x - 1=0$ gives $x = 1$. The x - intercepts are $(1,0)$ and $(3,0)$.

Step4: Find x - coordinate of vertex

For a quadratic function in standard form $y=ax^{2}+bx + c$ (here $a = 1$, $b=-4$, $c = 3$), the x - coordinate of the vertex is $x=-\frac{b}{2a}$. Substitute $a = 1$ and $b=-4$ into the formula: $x=-\frac{-4}{2\times1}=\frac{4}{2}=2$.

Step5: Find y - coordinate of vertex

Substitute $x = 2$ into $f(x)=x^{2}-4x + 3$. Then $f(2)=2^{2}-4\times2+3=4 - 8 + 3=-1$.

Answer:

a) $f(x)=x^{2}-4x + 3$
b) $(0,3)$
c) $(1,0)$; $(3,0)$
d) $2$
e) $-1$