19 (a) Sketch the graph of $y = (x - 2)^2 - 3$ - OCR - GCSE Maths - Question 19 - 2017 - Paper 1

From the graph provided, we can observe that the parabola opens upwards and the vertex lies at approximately $(-1, 12)$.

Using the vertex form of a quadratic equation $y = a(x - h)^2 + k$, we can identify that:

• The vertex $(h, k) = (-1, 12)$. Thus, $h = -1$ and $k = 12$.
• This gives us the equation:
  $y = a(x + 1)^2 + 12$

To find $a$, we also know the graph passes through the point $(0, y)$. Observing the graph, it seems to pass through $(0, 12)$:

1. Substitute $(0, y)$ into the equation: $12 = a(0 + 1)^2 + 12$ $12 = a(1) + 12$ $0 = a$

Since the opening of the parabola suggests it's wider, we may consider a = rac{1}{4}. Using another point from the graph $(3, 0)$: 2. Substitute $(3, y)$ into the equation: $0 = a(3 + 1)^2 + 12$ $0 = 16a + 12$ $16a = -12$ a = -rac{3}{4}

Solving $b$ and $c$ using $y = ax^2 + bx + c$: By equating the expanded form, we can find:

1. y = -rac{3}{4}(x^2 + 2x + 1) + 12 leads us to identify a = -rac{3}{4}, b = -rac{3}{2}, c = 0.

Thus, the final values are:

• a = -rac{3}{4},
• b = -rac{3}{2},
• $c = 0$.