Complete this table for $y = x^2 + x - 4$ - OCR - GCSE Maths - Question 5 - 2017 - Paper 1

To draw the graph of the function, plot the points from the completed table on the Cartesian plane. Then, connect the points with a smooth curve, depicting the quadratic nature of the function. The graph should be a parabola opening upwards. Make sure the x-axis covers the range from -4 to 3, and the y-axis appropriately reflects the range of values for $y$.

To solve for $x$ in the equation $x^2 + x - 4 = 0$, identify where the graph of $y = x^2 + x - 4$ intersects the x-axis. The x-coordinates of these points of intersection are the solutions to the equation. Based on the graph, estimate the values of $x$ for the intersections.

To draw the graph of $y = -2x - 1$, we can choose x-values from the specified range and calculate corresponding y-values:

• For $x = -4$:
  $y = -2(-4) - 1 = 8 - 1 = 7$
  So, the point is $(-4, 7)$.

• For $x = -3$:
  $y = -2(-3) - 1 = 6 - 1 = 5$
  Point is $(-3, 5)$.

• For $x = -2$:
  $y = -2(-2) - 1 = 4 - 1 = 3$
  Point is $(-2, 3)$.

• For $x = -1$:
  $y = -2(-1) - 1 = 2 - 1 = 1$
  Point is $(-1, 1)$.

• For $x = 0$:
  $y = -2(0) - 1 = -1$
  Point is $(0, -1)$.

• For $x = 1$:
  $y = -2(1) - 1 = -2 - 1 = -3$
  Point is $(1, -3)$.

• For $x = 2$:
  $y = -2(2) - 1 = -4 - 1 = -5$
  Point is $(2, -5)$.

• For $x = 3$:
  $y = -2(3) - 1 = -6 - 1 = -7$
  Point is $(3, -7)$.

Plot these points on the same grid as the first graph and draw a straight line through them.

To solve the equation, we need to rewrite it in the form that equals zero: $x^2 + x - 4 + 2x + 1 = 0 \Rightarrow x^2 + 3x - 3 = 0$

Next, we will find where the graphs of $y = x^2 + x - 4$ and $y = -2x - 1$ intersect. By finding the x-coordinates of these intersection points on the graph, we will obtain the solutions to the equation.