Question

Find the equation of the quadratic function whose graph is a parabola containing the points left parenthesis 0 comma 1 right parenthesis​, left parenthesis 4 comma 17 right parenthesis​, and left parenthesis negative 1 comma 7 right parenthesis.
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Part 1
The equation of the quadratic function whose graph is a parabola containing the given points is y equals nothing .
  

Answer 1

To find the equation of the quadratic function in the form \( y = ax^2 + bx + c \) that passes through the points \( (0, 1) \), \( (4, 17) \), and \( (-1, 7) \), we can set up a system of equations using these points.

1. From the point \( (0, 1) \):
\[
1 = a(0)^2 + b(0) + c \implies c = 1
\]

2. From the point \( (4, 17) \):
\[
17 = a(4)^2 + b(4) + c \implies 17 = 16a + 4b + 1 \implies 16a + 4b = 16 \implies 4a + b = 4 \quad (1)
\]

3. From the point \( (-1, 7) \):
\[
7 = a(-1)^2 + b(-1) + c \implies 7 = a - b + 1 \implies a - b = 6 \quad (2)
\]

Now we have a system of two equations:

1. \( 4a + b = 4 \) (Equation 1)
2. \( a - b = 6 \) (Equation 2)

Next, we can solve this system of equations. From Equation (2), we can express \( b \) in terms of \( a \):

\[
b = a - 6
\]

Now substitute \( b \) into Equation (1):

\[
4a + (a - 6) = 4
\]
\[
4a + a - 6 = 4
\]
\[
5a - 6 = 4
\]
\[
5a = 10 \implies a = 2
\]

Now, substitute \( a = 2 \) back into Equation (2) to find \( b \):

\[
2 - b = 6 \implies -b = 4 \implies b = -4
\]

Now we have \( a = 2 \), \( b = -4 \), and \( c = 1 \). Thus, the equation of the quadratic function is:

\[
y = 2x^2 - 4x + 1
\]

**Final answer:**
\[
y = 2x^2 - 4x + 1
\]