Question

identify the vertex of the quadratic function in standard form. remember to use x = -b/2a

1. y = 2x² - 16x + 31
2. y=-x² - 4x + 1
3. y = 3x² - 6x + 4

Explanation:

Step1: Recall vertex - x formula

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

Step2: For $y = 2x^{2}-16x + 31$

Here $a = 2$, $b=-16$. Then $x=-\frac{-16}{2\times2}=\frac{16}{4}=4$.

Step3: Find y - coordinate

Substitute $x = 4$ into $y = 2x^{2}-16x + 31$. So $y=2\times4^{2}-16\times4 + 31=2\times16-64 + 31=32-64 + 31=-1$. The vertex is $(4,-1)$.

Step4: For $y=-x^{2}-4x + 1$

Here $a=-1$, $b=-4$. Then $x=-\frac{-4}{2\times(-1)}=\frac{4}{-2}=-2$.

Step5: Find y - coordinate

Substitute $x=-2$ into $y=-x^{2}-4x + 1$. So $y=-(-2)^{2}-4\times(-2)+1=-4 + 8 + 1=5$. The vertex is $(-2,5)$.

Step6: For $y = 3x^{2}-6x + 4$

Here $a = 3$, $b=-6$. Then $x=-\frac{-6}{2\times3}=\frac{6}{6}=1$.

Step7: Find y - coordinate

Substitute $x = 1$ into $y = 3x^{2}-6x + 4$. So $y=3\times1^{2}-6\times1 + 4=3-6 + 4=1$. The vertex is $(1,1)$.

Answer:

1. $(4,-1)$
2. $(-2,5)$
3. $(1,1)$