I think the first one is 4x=3
Solve the differential equation dy dx equals the quotient of y times the cosine of x and the quantity 1 plus y squared with the
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The shape of the graph of quadratic functions is the shape of a parabola.
The steps for drawing a graph of the function f(x) = 2·(x + 4)² - 3 arranged in the correct order are;
- (a) Plot the vertex (-4, -3)
- (b) Apply the 2·x² pattern, from the vertex go right 1 up 2 and plot (-3, -1)
- (c) Apply the 2·x² pattern again, from the vertex go right 2 up 8 and plot (-2, 5)
- (d) Apply the 2·x² pattern a 3rd time, from the vertex go right 3 up 18 and plot (-1, 15)
- (e) Draw the axis of symmetry x = -4
- (f) Reflect the points over the axis of symmetry and plot (-5, -1), (-6, 5), and (-7, 15)
- (g) Draw a U shaped parabola.
Reasons:
The given function is; f(x) = 2·(x + 4)² - 3
The function is given in the vertex form; f(x) = a·(x - h)² + k
Therefore, the vertex, (h, k) = (-4, -3)
Step (a);
The vertex can be plotted on the graph
- Plot the vertex (-4, -3)
Step (b);
Given that the quadratic term is 2·x², the pattern that can be used for the points from the vertex is therefore, 2·x²
From the vertex (-4, -3) apply the 2·x² pattern by going to the right 1 unit and up 2 × 1² = 2 units to get the point (-4 + 1, -3 + 2) = (-3. -1)
- Apply the 2·x² pattern, from the vertex go right 1 up 2 and plot (-3, -1)
Step (c);
To get the next point, the 2·x² pattern is applied with x = 2, to the vertex to get; (-4 + 2, -3 + (2×2²)) = (-2, 5)
- Apply the 2·x² pattern again, from the vertex go right 2 up 8 and plot (-2, 5)
Step (d);
A third point on the graph relative to the vertex is obtained again by applying the 2·x² pattern again to the vertex with x = 3, to get;
(-4 + 3, -3 + (2 × 3²)) = (-1, 15)
- Apply the 2·x² pattern a 3rd time, from the vertex go right 3 up 18 and plot (-1, 15)
Step (e);
The axis of symmetry can be drawn with a vertical line passing through the vertex, which is the line, x = -4
The line x = -4 can be drawn on the graph next
- Draw the axis of symmetry x = -4
Step (f);
The points obtained relative to the vertex (-3, -1), (-2, 5), (-1, 15) can reflected about the axis of symmetry x = -4, to get;
(-3, -1) (-5, -1)
(-2, 5) (-7, 5)
(-1, 15) (-7, 15)
- Reflect the points over the axis of symmetry and plot (-5, -1), (-6, 5), and (-7, 15)
Step (g);
The parabola that is U shaped can be drawn from the points plotted in the steps above.
- Draw a U shaped parabola
Learn more about drawing the graph of a quadratic function here: