The graph of the quadratic function f(z) = ar²+bx+c has its vertex at the point A (1,7) and crosses the y-axis at the point B (0,2).

(a) Write down:

(1) the equation of the axis of symmetry.

(ii) the value of c.

(b) Find the value of a and the value of 6.

Given the quadratic function \( f(z) = az^2 + bz + c \), the vertex at point \( A(1, 7) \) and the point where it crosses the y-axis at \( B(0, 2) \), we can now answer the questions step-by-step:

### (a) 

#### (i) The equation of the axis of symmetry:

The axis of symmetry for a quadratic function in the form \( f(z) = az^2 + bz + c \) is given by the equation:
\[
z = -\frac{b}{2a}
\]
However, since we know the vertex is at \( z = 1 \), the axis of symmetry must be:
\[
z = 1
\]

#### (ii) The value of \( c \):

Since the point \( B(0, 2) \) represents where the graph crosses the y-axis, this means when \( z = 0 \), \( f(0) = 2 \). In the quadratic function, \( f(0) = c \), so:
\[
c = 2
\]

### (b) Finding the value of \( a \) and \( b \):

We know that the vertex \( A(1, 7) \) gives the equation of the function at \( z = 1 \), meaning \( f(1) = 7 \). Plugging this into the quadratic form:
\[
f(1) = a(1)^2 + b(1) + 2 = 7
\]
Simplifying:
\[
a + b + 2 = 7
\]
\[
a + b = 5 \quad \text{(Equation 1)}
\]

Additionally, we know the function passes through \( B(0, 2) \), which confirms that \( c = 2 \). Now, to find \( a \) and \( b \), we need one more piece of information. Since the axis of symmetry is \( z = 1 \), this gives us a relationship between \(

The graph of the quadratic function f(z) = az² + bz + c has its vertex at the point A (1,7) and crosses the y-axis at the point B (0,2). (a) Write down: (1) the equation of the axis of symmetry. (ii) the value of c. (b) Find the value of a and the value of b.

Solve a quadratic function with a vertex at A(1,7) and a y-intercept at B(0,2). Learn how to determine the axis of symmetry, the value of c, and how to calculate the coefficients a and b. Suitable for students in Grades 9-10.

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