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On the axes below sketch the graph of y = 4x² + 8x +3

Label all points of intersection and the turning point in your sketch.

On the axes below sketch the graph of y = 4x² + 8x +3 Label all points of intersection-example-1
User Jacob Ford
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1 Answer

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1. Find the vertex (turning point) of the parabola by using the formula x = -b/2a, where a = 4 and b = 8. This gives x = -1, which is the x-coordinate of the vertex. To find the y-coordinate, substitute x = -1 into the equation: y = 4(-1)² + 8(-1) + 3 = -1.

2. Plot the vertex at the point (-1, -1).

3. To find the x-intercepts, set y = 0 in the equation and solve for x. This gives x = (-8 ± √(8² - 4(4)(3)))/(2(4)) = (-8 ± √16)/8 = -1/2 or -3. Plot these points on the x-axis.

4. To find the y-intercept, set x = 0 in the equation and solve for y. This gives y = 3, so the y-intercept is at the point (0, 3).

5. Since the coefficient of x² is positive, the parabola opens upwards. Sketch the curve passing through the points found above.

6. Label the points of intersection and the turning point on the graph.

User Mykola Shorobura
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