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.

Explanation:

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 \( x =-\frac{0}{2\times(-4)}=0 \). Substitute \( x = 0 \) into the function \( y=-4x^{2}+7 \), we get \( y=-4\times0^{2}+7 = 7 \). So the vertex is \( (0,7) \).

Step2: Find points to the left of the vertex

Let's choose \( x=-1 \) and \( x = - 2 \) (left of \( x = 0 \)).

• 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 \( y=-4x^{2}+7 \), \( y=-4\times(-2)^{2}+7=-4\times4 + 7=-16 + 7=-9 \). So the point is \( (-2,-9) \).

Step3: Find points to the right of the vertex

Let's choose \( x = 1 \) and \( x=2 \) (right of \( x = 0 \)).

• For \( x = 1 \): Substitute into \( y=-4x^{2}+7 \), \( y=-4\times1^{2}+7=-4 + 7=3 \). So the point is \( (1,3) \).
• For \( x=2 \): Substitute into \( y=-4x^{2}+7 \), \( y=-4\times2^{2}+7=-4\times4+7=-16 + 7=-9 \). So the point is \( (2,-9) \).

Answer:

The five points are the vertex \((0,7)\), points to the left \((-1,3)\), \((-2,-9)\) and points to the right \((1,3)\), \((2,-9)\)