QUESTION IMAGE
Question
y = -4x² + 7
plot five points on the parabola: the vertex, two points to the left of the vertex, and two points to the right of the vertex. then click on the graph-a-function button.
Step1: Find the vertex
The quadratic function is in the form \( y = ax^2 + bx + c \), here \( a=-4 \), \( b = 0 \), \( c = 7 \). The x - coordinate of the vertex of a parabola \( y=ax^{2}+bx + c\) is given by \( x=-\frac{b}{2a} \). Substituting \( b = 0 \) and \( a=-4 \), we get \( x = 0 \). Substitute \( x = 0 \) into \( y=-4x^{2}+7 \), we get \( y=7 \). So the vertex is \( (0,7) \).
Step2: Find points to the left of the vertex (x negative)
Let's choose \( x=-1 \) and \( x = - 2 \).
- For \( x=-1 \): Substitute into \( y=-4x^{2}+7 \), \( y=-4\times(-1)^{2}+7=-4 + 7=3 \). So the point is \( (-1,3) \).
- For \( x=-2 \): Substitute into \( y=-4x^{2}+7 \), \( y=-4\times(-2)^{2}+7=-16 + 7=-9 \). So the point is \( (-2,-9) \).
Step3: Find points to the right of the vertex (x positive)
Let's choose \( x = 1 \) and \( x=2 \).
- For \( x = 1 \): Substitute into \( y=-4x^{2}+7 \), \( y=-4\times(1)^{2}+7=-4 + 7=3 \). So the point is \( (1,3) \).
- For \( x = 2 \): Substitute into \( y=-4x^{2}+7 \), \( y=-4\times(2)^{2}+7=-16 + 7=-9 \). So the point is \( (2,-9) \).
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The five points are \((0,7)\), \((-1,3)\), \((-2,-9)\), \((1,3)\), \((2,-9)\)